Re: [algogeeks] Maximize the Time to see TV

[Divya Jain]

@ above

ur code fails for following example
channel 1 : prog1 8:00-9:00, prog 2 9:00-10:00
channel 2 : prog1 8:15-10:00

your code returns 8:15- 10
and the answer should be channel1/prog1 + channel1/prog2



On 21 January 2011 12:54, Anand  wrote:

>
> Sort all program with their starting time.
>
> Appy the below pseudo code to find max number of programs he can watch.
>
> for(i=i;i {
> /*Check for overlap */
>if(p[i].start > p[i-1].end)
>{
> end = i;
>}
>else
>{
>   /*Index of the first program to be watch*/
>   if((p[i-1].end - p[i-1].start) < (p[i].end - p[i].start))
>   {
>  start = i;
>   }
>}
>
> }
>  return end - start;
>
>
> On Thu, Jan 20, 2011 at 10:11 PM, snehal jain wrote:
>
>> There is a TV avid person. HE wants to spend his max time on TV. There
>> are N channels with different program of different length and diff
>> times. WAP so that the person cam spend his max time watching TV.
>> Precondition: If that person watches a program, he watches it
>> completely.
>>
>> Ex:
>> Channel1: prog1 – 8:00- 8:30
>> prog2: 9:00 – 10:00
>> prog3: 10:15 – 12:00
>>
>> channel2: prg1 – 8:15 – 10:00
>> prg2: 10:30 – 12:00
>>
>> So in this case max time will be if he watches:
>>
>> ch2/prg1 + ch1/prg3
>>
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Re: [algogeeks] find a way

[Divya Jain]

if u can take only a certain amount of fuel from a particaular station ie
station xi can provide li amoutn of fuel.. then wat?

On 22 January 2011 13:46, Terence  wrote:

> Greedy-Approach.
> Refueling only when you have to.
>
>
>
> On 2011-1-22 15:59, snehal jain wrote:
>
>> Suppose you want to travel from city A to city B by car, following a
>> fixed route. Your car can travel m miles on a full tank of gas, and
>> you have a map of the n gas stations on the route between A and B that
>> gives the number of miles between each station.
>> Design an algorithm to find a way to get from A to B without running
>> out of gas that minimizes the total number of refueling stops, in O(n)
>> time.
>>
>>
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Re: [algogeeks] Re: find a line

[Divya Jain]

assume the region to be (0,0) , (10,0), (0,10), (10,10)

On 22 January 2011 08:33, Dave  wrote:

> @Divya: The coordinates of the points are between 0 and 1 and are
> integers. That can't be right.
>
> Dave
>
> On Jan 21, 1:46 pm, divya  wrote:
> > Within a 2D space, there is a batch of points(no duplicate) in the
> > region (0,0),(0,1),(1,0),(1,1), try to find a line which can divide
> > the region to 2 parts with half points in each .the input will be an
> > array of points and the length of the array.
> > struct point{
> > int x;
> > int y;};
> >
> > input : struct point * points, int length
>
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Re: [algogeeks] Adobe Question

[Divya Jain]

if both arrray are sorted. take two ptrs, one pointing to a[0] other to
b[0]. if elements are same then nt disjoint. else increment ptr pointing to
smaller element until it becomes equal or greater than the element pointed
by other. repeat until either ptr reaches end of array..

On 21 January 2011 21:42, Manmeet Singh  wrote:

> how merge sort ?/
>
>
> On Thu, Jan 20, 2011 at 11:23 PM, nishaanth  wrote:
>
>> Ya as Ashish said hashing is the best solution :)
>>
>>
>> On Fri, Jan 14, 2011 at 6:00 PM, Ashish Goel  wrote:
>>
>>> ideally, a hashMap would be preferred
>>>
>>> walk through one array and set the corresponding entry, and then through
>>> another array, if any entry found, then they are not disjoint.
>>>
>>> Best Regards
>>> Ashish Goel
>>> "Think positive and find fuel in failure"
>>> +919985813081
>>> +919966006652
>>>
>>>
>>> On Fri, Jan 14, 2011 at 3:35 PM, bittu wrote:
>>>
 how to find if two arrays of size n are disjoint or not in O(n)
 time ?? You can use only O(n) space
 The elements are +ve in the range 0 to n power 100..



 Regards
 Shashank Mani

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>> Computer Science and engineering,
>> IIT Madras.
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Re: [algogeeks] merging 2 search trees

[Divya Jain]

@ above height will not be balanced then

On 21 January 2011 19:15, nishaanth  wrote:

> Find the node in T which is the maximum(which is either the root or the
> rightmost in the right subtree).
> After finding this node, just make the right child of this node point to
> the root of T'.
>
> Correct me if i am wrong
>
>
> On Fri, Jan 21, 2011 at 2:43 PM, snehal jain wrote:
>
>> You are given two height balanced binary search trees T and T’,
>> storing m and n elements respectively. Every element of tree T is
>> smaller than every element of tree T’. Every node u also stores height
>> of the subtree rooted at it. Using this extra information how can you
>> merge the two trees in time O(log m + log n) (preserving both the
>> height balance and the order)?
>>
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>
>
> --
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> Computer Science and engineering,
> IIT Madras.
>
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Re: [algogeeks] Adobe Question

[Divya Jain]

if sorted then a tweak in merge sort will work

On 20 January 2011 23:23, nishaanth  wrote:

> Ya as Ashish said hashing is the best solution :)
>
>
> On Fri, Jan 14, 2011 at 6:00 PM, Ashish Goel  wrote:
>
>> ideally, a hashMap would be preferred
>>
>> walk through one array and set the corresponding entry, and then through
>> another array, if any entry found, then they are not disjoint.
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>> On Fri, Jan 14, 2011 at 3:35 PM, bittu wrote:
>>
>>> how to find if two arrays of size n are disjoint or not in O(n)
>>> time ?? You can use only O(n) space
>>> The elements are +ve in the range 0 to n power 100..
>>>
>>>
>>>
>>> Regards
>>> Shashank Mani
>>>
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>
>
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Re: [algogeeks] Re: tic tac toe

[Divya Jain]

@above yes
@nishaanth that will b n^2

On 9 January 2011 17:45, juver++  wrote:

> What do you mean by N term? Is it a size of the matrix N*N?
>
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Re: [algogeeks] Re: difference x

[Divya Jain]

its okk

On 22 December 2010 22:28, Ankur Khurana  wrote:

> saurabh , asking for a better sol. is not a crime. rest everybody is
> intelligent.
>
> On Wed, Dec 22, 2010 at 10:27 PM, Saurabh Koar 
> wrote:
> > @Snehal: I hv no problem wid ur doubts.Bt sometimes u post probs which
> > r very basic.As u solved the looping prob in other thread I think u r
> > intelligent enough to solve this prob(difference x) easily and also
> > some other probs posted by u(not all).Some problems posted by u are
> > really very tricky and I loved them.I jst requested u while posting if
> > u review the ques carefully it will be helpful.I dint want to hurt
> > u.If u r I m sorry.Bt still I will request u to post carefully.Lets
> > finish this.And sorry once again.
> >
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Re: [algogeeks] Re: difference x

[Divya Jain]

@ saurabh
and i wanna ur solution to this prob
i know O(nlogn) post ur O(n) solution,.
i hv some idea bt i wanna knw ur ans b4 mine

On 22 December 2010 22:32, Divya Jain  wrote:

> its okk
>
>
> On 22 December 2010 22:28, Ankur Khurana  wrote:
>
>> saurabh , asking for a better sol. is not a crime. rest everybody is
>> intelligent.
>>
>> On Wed, Dec 22, 2010 at 10:27 PM, Saurabh Koar 
>> wrote:
>> > @Snehal: I hv no problem wid ur doubts.Bt sometimes u post probs which
>> > r very basic.As u solved the looping prob in other thread I think u r
>> > intelligent enough to solve this prob(difference x) easily and also
>> > some other probs posted by u(not all).Some problems posted by u are
>> > really very tricky and I loved them.I jst requested u while posting if
>> > u review the ques carefully it will be helpful.I dint want to hurt
>> > u.If u r I m sorry.Bt still I will request u to post carefully.Lets
>> > finish this.And sorry once again.
>> >
>> > --
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>> Groups "Algorithm Geeks" group.
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Re: [algogeeks] linkd list

[Divya Jain]

no not sorted

On 15 December 2010 11:37, Anand  wrote:

> @Divya, I think you ask this question before. Not sure of the answer though
> :)
>
>
> On Tue, Dec 14, 2010 at 9:06 PM, Soumya Prasad Ukil  > wrote:
>
>> Are the linked list sorted?
>>
>>
>> On 14 December 2010 05:33, divya  wrote:
>>
>>> Given three lists: A, B and C of length n each. Write a function which
>>> returns true if any 3 three numbers (1 from each list), sum up to
>>> zero. Else return false, if there is no such combination.
>>>
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Re: [algogeeks] Re: how many bananas?

[Divya Jain]

nice solution:)

On 14 December 2010 20:54, Prims  wrote:

> total bananas=3000 ...max capacity=1000
> ==>3 trips needed to transport all
>
> for traveling 1 km
> 1st trip 998[eats 1 banana while coming and 1 banana while going back]
> 2nd trip 998
> 3rd trip 999[no need to go back]
>
> ==>5 banana for travelling 1 km
>
> since 3 trips involved v spend 5 banana per km
> v need to reduce it to 2 trips
> ==>the point where total banana becomes 2000
> ==>3000-2000=1000
> ==>1000/5=200 km
>
> at 200 km v hav 2000 bananas
> now for travelling 1 km
> 1st trip 998
> 2nd trip 999
>
> ==>3 banana for travellin 1 km
>
> find the point where this becomes single trip
> dat is total banana becomes 1000
> ==>2000-1000=1000
> ==>1000/3=333.3km
>
> v hav reached 533.33 km with 1000 banana
> now make a single trip of remaining 466.66 to city B
> v reach city B with 533.3 banana
>
> dats the answer 533 1/3 banana
>
>
> On Dec 14, 5:47 pm, divya  wrote:
> > There is a desert of 1000kms long. A camel is there and 3000 bananas.
> > At one go a camel can take 1000 bananas. For each one km to walk camel
> > eats 1 banana. How many bananas can we cross to the other side of
> > desert
>
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Re: [algogeeks] Re: sorted array

[Divya Jain]

ignore the above post of mine
@ shiyam
u r doing wrong. we hv been the size of both array b and c which is 5 n 3
respectively

@ AlgoSAu
can u please explain ur method?

On 13 October 2010 17:23, Divya Jain  wrote:

> @above
> can u plz explain hw did u apply this formula?
>
>
> On 13 October 2010 16:14, Shiyam code_for_life wrote:
>
>> Lets say 8 integers are 1 to 8
>>
>> So first way array can be split is 1,7
>>
>> 1st array can have one among 8 integers so in 8 ways and 2nd array has
>> just one arrangement of other integers in sorted manner and same goes
>> on for other ways to split the array.
>>
>> Whats wrong here?
>>
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Re: [algogeeks] Re: sorted array

[Divya Jain]

@above
can u plz explain hw did u apply this formula?

On 13 October 2010 16:14, Shiyam code_for_life wrote:

> Lets say 8 integers are 1 to 8
>
> So first way array can be split is 1,7
>
> 1st array can have one among 8 integers so in 8 ways and 2nd array has
> just one arrangement of other integers in sorted manner and same goes
> on for other ways to split the array.
>
> Whats wrong here?
>
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Re: [algogeeks] Maximum size binary search tree

[Divya Jain]

@ above
ur soln ll fail in situation like
  10
 / \
  15   18
 /\  /  \
22   7  17  77
the inorder is
22 15 7 10 17 18 77
so the longest increasing sequence is 7-77
but this is not a bst as 10 n 7 r nt connected

On 24 June 2010 22:36, gopinath sikha  wrote:

> We can find the solution in O(n) where n is number of nodes.
> Do an in-order traversal of the binary tree. then scan through the numbers
> and find the list and find the longest(increasing or decreasing) sequence.
> That is the size of maximum size of BST in the given binary tree.
>
>
> On Wed, Jun 23, 2010 at 11:40 AM, Raj N  wrote:
>
>> Find the maximum size Binary search tree in a binary tree??
>>
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Re: [algogeeks] Array with 0 and 1

[divya jain]

i think u need to visit every element atleast once to see if its 1 or 0, nd
so update the count.
so i dont think it will be possible in less than O(n2)

On 5 July 2010 15:41, amit  wrote:

>
>
> For a given matrix NxN having 0 or 1’s only. You have to find the
> count of rows and columns where atleast one 1 occurs.
>
> e,g
>
> 0 0 0 0
>
> 1 0 0 1
>
> 1 0 0 1
>
> 1 1 0 1
>
> Row count having 1 atleast once: 3
>
> Col count having 1 atleast once: 3
>
> Any Solution less than O(n^2) will do
>
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Re: [algogeeks] amazon question

[divya jain]

serialize means u hv to store binary tree in an array in contiguous
locations ( not like the way we store tree in an array usually where
children are at 2i+1 n 2i+2 if parent is at i) such that u can reconstruct
tree later easily from the array and u shd get the same tree..

@ harit can u please elaborate approach.. wen r u putting ',' if u r putting
wen u visit leaf then it shd b there r after 10 too. m nt getting??

On 3 July 2010 19:42, ashish agarwal  wrote:

> We can do 4 type of treversal.
> If we do inorder then we will get sorted array .If we do an inorder
> traversal then we would get a sorted list which if we convert into a BST
> would again become a list and we would loose out on the original structure
> of the tree.
>
> and same will be happen with post order
> now remaining preorder and level order.
> when we do level order traversal it will require more space as it uses BFS
> approach .So to do in o(logn) we do preorder travesal.
>
>
> On Sat, Jul 3, 2010 at 5:46 AM, jalaj jaiswal 
> wrote:
>
>> serialize... is it level order traversal ???
>> give an example...?
>>
>>
>> On Sat, Jul 3, 2010 at 12:36 PM, divya  wrote:
>>
>>> Design an algorithm and write code to serialize a binary tree.
>>>
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Re: [algogeeks] Re: isbst

[divya jain]

@ above

here u r comparing node value with min and max only
for eg if ur tree is

  45
/  \
  65   85
/
  25

ur code will say that this is bst. bt it is not
plzz correct me if i m wrong..

On 1 July 2010 16:17, CM Saikanth Varma  wrote:

> Hi,
>
> The idea is like this:
>
> Isbst(node *t){
> if(t==NULL){ return true; }
> if (  Isbst(t->left) &&
>  Isbst(t->right) &&
>  (maxValue(t->left) <=(t->data) ) &&
>  (minValue(t->right) >= (t->data)
> )
>  return true;
> else
> return false;
> }
>
> I hope it makes sense :D
> On Thu, Jul 1, 2010 at 3:37 PM, vicky  wrote:
>
>> just do one thing print inorder. if sorted it is a BST
>>
>> On Jun 19, 2:57 pm, divya  wrote:
>> > i have found the following code on net to check if the binarytreeis
>> > bst or not
>> > /*
>> > Returns true if a binarytreeis a binary searchtree.
>> > */
>> > int isBST(struct node* node) {
>> > if (node==NULL) return(true);
>> > // false if the min of the left is > than us
>> > if (node->left!=NULL && minValue(node->left) > node->data)
>> > return(false);
>> > // false if the max of the right is <= than us
>> > if (node->right!=NULL && maxValue(node->right) <= node->data)
>> > return(false);
>> > // false if, recursively, the left or right is not a BST
>> > if (!isBST(node->left) || !isBST(node->right))
>> > return(false);
>> > // passing all that, it's a BST
>> > return(true);
>> >
>> > }
>> >
>> > int minValue(struct node* node) {
>> > struct node* current = node;
>> > // loop down to find the leftmost leaf
>> > while (current->left != NULL) {
>> > current = current->left;}
>> >
>> > return(current->data);
>> >
>> > }
>> >
>> > and maxvalue also similarly returns right most vaslue oftree..
>> >
>> > now i have a doubt that here v r comparing the node->data with only
>> > the min node nd max node.. shd nt we compare the node data with its
>> > left node and right node only.
>> > as it can b a case that node value is greater than min but less than
>> > its left node.. nd here we r nt checking that...
>> >
>> > plzz correct me if i m wrong...
>> >
>> > and is there any other efficient method to find isbst?
>>
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Re: [algogeeks] second shortest path

[divya jain]

my approach is like this
remove any 1 edge from the graph and now find shortest path b/w reqd pts
without this edge. store the shortest path in some variable
now restore the removed edge and remove some other edge and again find
shortest path. if this is shorter than prev stored value then update that
else ignore
do this for all edges ie remove all edge one by one at a time n compare the
shortest path with the value stored in the variable
at end that variable wil giv u shortest path
i think there must b more efficient soln
hope m clear this time

On 28 June 2010 23:11, sharad kumar  wrote:

> @divya
> plzzz elaborate ur approach a bit more...i didnt get u
>
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Re: [algogeeks] second shortest path

[divya jain]

keep on removing one edge at a time n find the shortest path..
the shortest path of all these paths is the ans..
On 27 June 2010 23:40, sharad kumar  wrote:

>  How to find the second shortest path in a graph?
>
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Re: [algogeeks] a bst

[divya jain]

@algoose
yes u r ryt

On 26 June 2010 18:01, Raj N  wrote:

> Sorry that's shud have been node->right==node->right->left
>
>
> On Sat, Jun 26, 2010 at 5:59 PM, Raj N  wrote:
>
>> #define max(x,y) ((x)>(y)?(x):(y))
>>
>> struct Bintree{
>>  int element;
>>  struct Bintree *left;
>>  struct Bintree *right;
>> };
>>
>> typedef struct Bintree* Tree;
>>
>> int height(Tree T)
>> {
>>  if(T->right==T->right->left)
>>  return -1;
>>  else
>>  return (1 + max(height(T->left), height(T->right)))
>> }
>>
>> Instead of checking the condition for node==NULL check if
>> node->right->left. This has got nothing to do with BST by the way..
>>
>>
>> On Sat, Jun 26, 2010 at 3:54 PM, divya jain wrote:
>>
>>> yes..
>>>
>>> i got the solution traverse till node->right!=node->right->left... at
>>> this point u ll get height.. rite?
>>>
>>> On 26 June 2010 11:49, Raj N  wrote:
>>>
>>>> @Divya: What will happen when say node->right when you reach the leaves
>>>> ? Is it equivalent to node->next and node->left = = node->previous in the
>>>> doubly linked list ?
>>>>
>>>>
>>>> On Tue, Jun 22, 2010 at 4:44 PM, divya wrote:
>>>>
>>>>> a bst is given whose leaf nodes having left as well as right pointers
>>>>> not pointing to NULL. rather all the leaf nodes are forming a circular
>>>>> doubly linked list. u have to calculate height of tree.
>>>>>
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Re: [algogeeks] a bst

[divya jain]

yes..

i got the solution traverse till node->right!=node->right->left... at this
point u ll get height.. rite?

On 26 June 2010 11:49, Raj N  wrote:

> @Divya: What will happen when say node->right when you reach the leaves ?
> Is it equivalent to node->next and node->left = = node->previous in the
> doubly linked list ?
>
>
> On Tue, Jun 22, 2010 at 4:44 PM, divya  wrote:
>
>> a bst is given whose leaf nodes having left as well as right pointers
>> not pointing to NULL. rather all the leaf nodes are forming a circular
>> doubly linked list. u have to calculate height of tree.
>>
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Re: [algogeeks] c

[divya jain]

ya it is giving error cz u r incrementing x. u cant increment x as it is
const. but there is no prob with declaration. its valid.

On 26 June 2010 11:09, sharad kumar  wrote:

> pls c thz
>
>
> On Sat, Jun 26, 2010 at 10:48 AM, sharad kumar wrote:
>
>> cos if u declare const u cant change na.but volatile changes na.so
>> practically u cant declareU cant do mtech at IIT and MBA at IIM at same
>> time rite???
>>
>>
>> On Sat, Jun 26, 2010 at 10:43 AM, sharad kumar 
>> wrote:
>>
>>> no its not cos u have declared const and volatile hw is it possible
>>> can u go to IIT and IIM at same time
>>>
>>>   On Sat, Jun 26, 2010 at 9:22 AM, sharad kumar <
>>> sharad20073...@gmail.com> wrote:
>>>
 static const volatile int x;
 Is the declaration valid
 plzzz explain

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>>>
>>>
>>>
>>> --
>>> yezhu malai vaasa venkataramana Govinda Govinda
>>>
>>
>>
>>
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Re: [algogeeks] Re: There is an array of odd and even numbers. Now, sort them in such a way that the top portion of the array contains odd numbers, bottom portion contains even numbers

[divya jain]

@ gurav vivek

i think u r only exchanging odd and even.. only putting even to end and odd
in front and not doing descending sorting on odds and ascending on even..

plz correct me if i hv missed something..

On 24 June 2010 22:49, Vivek Sundararajan  wrote:

> Hi, how about this -
>
> Do a merge sort, now, while merging two sorted list, give more priority to
> odd numbers :)
>
> I believe this falls into the right solutions :)
>
> Any breaking cases?
>
>
> On 24 June 2010 09:41, Gaurav Singh  wrote:
>
>> I think in this case, bubble sorting will be a better idea. just
>> replace the condition of comparison with the condition that earlier
>> number is even and later number is odd. I mean we can do sumthing lyk
>> this :
>>
>> for i=1 to n-1
>>  for j=1 to n-i-1
>> if iseven(ar[j]) AND (NOT iseven(ar[j+1]))
>> then   swap both of them.
>>
>> Please correct me if I am wrong somewhere.
>>
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Re: [algogeeks] sum k in sub array

[divya jain]

@amir

ur algo works only for positive elements array..

correct me if m wrong

On 23 June 2010 00:28, Amir hossein Shahriari <
amir.hossein.shahri...@gmail.com> wrote:

> for each element find sum[i] which is the summation of all elements with
> index less than or equal to i ( note that this array is always sorted) this
> can be done in O(n)
> then sum[i]-sum[j] when j then for each sum[i] binary search for sum[i]-k in the array sum
> which yields the overall running time of O(nlogn)
>
> On Tue, Jun 22, 2010 at 7:48 PM, sharad kumar wrote:
>
>> How will you find the subarray whose sum is k in an array of negative and
>> positive numbers
>>
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Re: [algogeeks] linked list

[divya jain]

if we dont want to change original list..
is there ny efficient method for that?

On 23 June 2010 06:49, ANUJ KUMAR  wrote:

> reverse one of the linked lists in O(n) time and then see if the
> resulting one is same as the other one
>
> On Wed, Jun 23, 2010 at 1:56 AM, divya  wrote:
> > Two link lists are given ...check if one is reverse of other. Do it
> > without using extra space.
> >
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Re: [algogeeks] Re: no of bits

[divya jain]

@ dave
how did u come to this formula... m nt getting the logic..

@ sathaiah
yes u r rite

On 22 June 2010 19:32, chitta koushik  wrote:

> For more such problems and solns
>
> http://graphics.stanford.edu/~seander/bithacks.html
>
>
> for (c = 0; v; c++)
> {
>   v &= v - 1; // clear the least significant bit set
> }
>
> O(k)   -- no. of bits set in the number
>
>
>
> --Koushik C
>
>
> On Tue, Jun 22, 2010 at 7:16 PM, Dave  wrote:
>
>> Assuming 32 bit integers:
>> n = ((x >> 1) & 0x) + (x & 0x)
>> n = ((n >> 2) & 0x) + (n % 0x)
>> n = ((n >> 4) & 0x0F0F0F0F) + (n & 0x0F0F0F0F)
>> n = ((n >> 8) & 0x00FF00FF) + (n & 0x00FF00FF)
>> n = ((n >>16) & 0x) + (n & 0x)
>>
>> n now is the number of bits set in x. Time is O(1).
>>
>> Dave
>>
>> On Jun 22, 6:26 am, divya  wrote:
>> > find the no. of bits set in a no. in logn time
>>
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Re: [algogeeks] queue

[divya jain]

if u r linked queue then wud nt be a problem
if u r otherwise using array n u find queue is full then u can insert in
i+1th queue nd so on

On 20 June 2010 21:20, Piyush Verma <114piy...@gmail.com> wrote:

> @divya
> if ith queue is full and i want to enque some value in ith queue then what
> should be the condition
>
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Re: [algogeeks] queue

[divya jain]

there is a linked list where each node of linked list points to queue.. for
eg. 1st node points to queue 1 2nd to queue 2 nd so on..
hope m clear this time..

On 20 June 2010 18:30, Piyush Verma <114piy...@gmail.com> wrote:

> @divya
> plzz explain little more. m not getting...
>
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Re: [algogeeks] queue

[divya jain]

use a linked list where each node points to a queue

On 20 June 2010 15:13, sharad  wrote:

> Give a good data structure for having n queues ( n not fixed) in a
> finite memory segment. You can have some data-structure separate for
> each queue. Try to use at least 90% of the memory space.
>
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Re: [algogeeks] balance tree

[divya jain]

if there is no buffer then use avl insertion

On 20 June 2010 16:23, sharad kumar  wrote:

> @sharad:pls tell is there a buffer to store the  previous numbers.if yes
> then  then recursively aplly median finding theorem to sorted array ...
>
>
> On Sun, Jun 20, 2010 at 4:15 PM, sharad  wrote:
>
>> Given an incoming stream of sorted numbers construct a binary search
>> tree. At each stage, you should have a nearly balanced tree.
>>
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Re: [algogeeks] Re: non recursive tree traversals

[divya jain]

/*
   Assuming you have a stack setup with push() and pop() operations.
   Also assuming that all nodes are initially marked to 0.
   (This function will reset them back to zero when finished)
*/
void postorder(Node *n) {
  push(n);

  while (stack.size > 0) {
n = (Node*)pop();

if (n->marked || (n->left == NULL && n->right == NULL)) {
  n->marked = 0;
  printf("%d\n", n->value);
}
else {
  n->marked = 1;
  push(n);

  if (n->right) push(n->right);
  if (n->left) push(n->left);
}
  }
}

 is the above solution fine for postorder. plz let me knw if there is any
mistake

On 17 June 2010 10:37, Gene  wrote:

> On Jun 16, 3:01 pm, divya  wrote:
> > plz give algos of inorder, preorder nd post order tree traversal..non
> > recursive one..using stack..
> > nd the tree is not threaded
>
>
> #include 
> #include 
>
> typedef struct node_s {
>  int val;
>  struct node_s *left, *right;
> } NODE;
>
> NODE *new(int val)
> {
>  NODE *tree = malloc(sizeof(NODE));
>  tree->val = val;
>  tree->left = tree->right = NULL;
>  return tree;
> }
>
> void search(NODE *tree)
> {
>  if (tree) {
>printf("preorder %d\n", tree->val);
>search(tree->left);
>printf("inorder %d\n", tree->val);
>search(tree->right);
>printf("postorder %d\n", tree->val);
>  }
> }
>
> // Direct conversion of recursive code to iteration.
> struct stack_elt_s {
>  int site;
>  NODE *tree;
> } stack[100];
> int sp = 0;
>
> void search_iterative(NODE *tree) {
>  start:
>  if (tree) {
>printf("preorder %d\n", tree->val);
>// simulate the recursive call search(tree->left)
>stack[sp].tree = tree;
>stack[sp++].site = 0;
>tree = tree->left;
>goto start;
>  L0:
>printf("inorder %d\n", tree->val);
>// simulate the recursive call search(tree->right)
>stack[sp].tree = tree;
>stack[sp++].site = 1;
>tree = tree->right;
>goto start;
>  L1:
>printf("postorder %d\n", tree->val);
>  }
>  // simulate return to last call site
>  if (sp) {
>tree = stack[--sp].tree;
>switch (stack[sp].site) {
>case 0: goto L0;
>case 1: goto L1;
>}
>  }
> }
>
> int main(void)
> {
>  struct node_s *n0 = new(0);
>  struct node_s *n1 = new(1);
>  struct node_s *n2 = new(2);
>  struct node_s *n3 = new(3);
>  struct node_s *n4 = new(4);
>  n0->left = n1;
>  n0->right = n2;
>  n1->left = n3;
>  n2->left = n4;
>
>  printf("recusive:\n");
>  search(n0);
>
>  printf("\nnow iterative:\n");
>  search_iterative(n0);
>
>  return 0;
> }
>
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Re: [algogeeks] doubly linked list

[divya jain]

yes u need to start frm beg..


On 16 June 2010 17:15, Vivek Sundararajan  wrote:

> so, this means that i can traverse the list only from the beginning of the
> link list right?
>
> what if im given a pointer pointing to some node other than the head of the
> doubly linked list? will i be able to traverse in any direction now?
>
> please let me know if im missing something :)
>
> Thank you,
> Vivek
>
>
> On 16 June 2010 15:37, divya jain  wrote:
>
>> u can xor linked list. such that every node link contains the xor of its
>> prev nd next node address.. since for 1st node prev is null ( 0) so its link
>> contains only next. now to calculate next of 2nd node xor its link with 1st
>> node's link nd u ll get 3 rd node.s adddress nd so on..
>>
>> u can also use sum. store in link the sum of prev node n next node
>> address.. bt this cn result in overflow. so xor method is better
>>
>>
>> On 16 June 2010 09:14, sharad kumar  wrote:
>>
>>> u have to use XOR linked list
>>>
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Re: [algogeeks] sum to 0

[divya jain]

sort array c only.
now select every pair from array a nd b ( o(n2))
for every pair find the element from c which gives sum of the three element
0. ( search using binary search as c is sorted)
so the algo is O (n^2 logn))

On 16 June 2010 13:47, Amir hossein Shahriari <
amir.hossein.shahri...@gmail.com> wrote:

> @algoose chase: you're right thx for the test case
>
> On Wed, Jun 16, 2010 at 11:45 AM, Algoose Chase wrote:
>
>> @ Amir:
>> I think you cannot find two numbers in B and C such that their sum is
>> -A[k] in O(n) in all cases with the logic that you mentioned.
>>
>> for instance you can try with :
>> A : 2 7 8 10
>> B :-2, 0, 1, 9
>> C: -7 -2 1 3
>>
>> On Wed, Jun 16, 2010 at 5:56 AM, Amir hossein Shahriari <
>> amir.hossein.shahri...@gmail.com> wrote:
>>
>>> sort all arrays
>>> for each number in A , A[k]
>>> find two numbers in B and C such that their sum is -A[k]
>>> this can be done in O(n) time:
>>>
>>> set i at the beginning of B and j at the end of C
>>> while i=0
>>>   if ( B[i] + C[j] == -A[k] ) return true
>>>   if ( B[i] + C[j] < -A[k] ) increase i
>>>   else decrease j
>>>
>>> we have to do this for each element of A so the overall complexity is:
>>> O(n2) time O(1) space
>>>
>>> correct me if I'm wrong
>>>
>>> On 6/15/10, sharad kumar  wrote:
>>> > plzzz comment on this question
>>> >
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Re: [algogeeks] doubly linked list

[divya jain]

u can xor linked list. such that every node link contains the xor of its
prev nd next node address.. since for 1st node prev is null ( 0) so its link
contains only next. now to calculate next of 2nd node xor its link with 1st
node's link nd u ll get 3 rd node.s adddress nd so on..

u can also use sum. store in link the sum of prev node n next node address..
bt this cn result in overflow. so xor method is better

On 16 June 2010 09:14, sharad kumar  wrote:

> u have to use XOR linked list
>
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Re: [algogeeks] Re: file handing

[divya jain]

why a.c gets closed. well comma operator has left to right associativity.
for eg
a=(2,3,4) ;
here is a=4 after the statement is executed so similarly why not here. y c.c
is not closed here?

On 13 June 2010 22:16, souravsain  wrote:

> For Problem 3 see section "4.2 Using feof() incorrectly" in
> http://www.drpaulcarter.com/cs/common-c-errors.php
>
> On Jun 13, 8:36 pm, divya jain  wrote:
> > sorry i pasted wrong questn unser 2..
> >
> > the real question is
> > which file will get closed through fclose()
> > #include
> > int main()
> > {
> > FILE *fp,*fs,*ft;
> > fp=fopen("a.c","r");
> > fs=fopen("b.c","r");
> > ft=fopen("c.c","r");
> > fclose(fp,fs,ft);
> > return 0;
> >
> > }
> >
> > 3. yes it is feof..srry typed it wrong... nd fgets(str,80,fp) is
> perfectly
> > fine.. now the ans to this questn is that last line of the file will be
> > printed twice...( which i m unable to get why)...plzz explain...
> >
> > @ souravsain plzz ignore this mail..srry for the inconvenience..
> >
> > On 13 June 2010 17:37, jalaj jaiswal  wrote:
> >
> >
> >
> > > in question 1... ch gets the value of EOF... so first kicit 44-a
> > > gokulpeth\0 nagpur will get printed and then the value of EOF..
> >
> > >  question number 2 .. seems to me as nrml ...i think myfile.c only gets
> > > closed
> >
> > > in question number 3..it shld be fgets(str,79,fp)
> >
> > > On Sun, Jun 13, 2010 at 2:49 PM, divya 
> wrote:
> >
> > >> 1. wat ll be the o/p. plz explain y?
> > >> // abc.c contains "kicit 44-a gokulpeth\0 nagpur"
> > >> #include
> > >>  #include
> > >>  int main()
> > >>  {
> > >>  unsigned char ch;
> > >>  FILE *fp;
> > >>  fp=fopen("abc.c","r");
> > >>  if(fp==NULL)
> > >>  {
> > >>  printf("unable to open the file \n");
> > >>  exit(1);
> > >>  }
> > >>  while((ch=getc(fp))!=EOF)
> > >>  printf("%c",ch);
> > >>  fclose(fp);
> > >>  printf("\n",ch);
> > >>  return 0;
> > >>  }
> >
> > >>  2.which file will get closed through fclose() in the following
> > >> program and why?
> > >> #include
> > >>  int main()
> > >>  {FILE *fp;
> > >>  char ch;
> > >>  int i=1;
> > >>  fp=fopen(myfile.c","r");
> > >>  while((ch=getc(fp)!=EOF))
> > >>  {
> > >>  if(ch=='\n')
> > >>  i++;
> > >>  }
> >
> > >>  fclose(fp);
> > >>  return 0;
> > >>  }
> >
> > >>  3.point out the error if any in following
> >
> > >> #include
> > >>  int main()
> > >>  {
> > >>  FILE *fp;
> > >>  char str[80];
> > >>  fp=fopen("trial","r");
> > >>  while(!eof(fp))
> > >>  {
> > >>  fgets(str,80,fp);
> > >>  puts(str);
> > >>  }
> > >>  fclose(fp);
> > >>  return 0;
> > >>  }
> >
> > >> --
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> >
> > > --
> > > With Regards,
> > > Jalaj Jaiswal
> > > +919026283397
> > > B.TECH IT
> > > IIIT ALLAHABAD
> >
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Re: [algogeeks] trees

[divya jain]

reverse in order traversal till u reach kth node. reverse inorder means
first visit right child then print data nd then left.

On 14 June 2010 08:34, Lekha  wrote:

> Inorder traversal till u reach the kth element(If it is sorted in
> descending order, otherwise go till (n-k)th element)..
>
>
> On Sun, Jun 13, 2010 at 9:24 PM, Rohit Saraf 
> wrote:
>
>> Repeated q. Search in the group
>>
>> --
>> --
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>> Dept. of Computer Science and Engineering
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>> http://www.cse.iitb.ac.in/~rohitfeb14
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Re: [algogeeks] Re: Array Problem

[divya jain]

i meant make an array of indexes.. index[]={0,1...,n-1}

now sort the original array and move the elements of index array
accordingly..

now follow modelling expert's algo nd while taking largest-smallest check if
the index of largest in index array > index of smallest in index array.

On 13 June 2010 08:38, Rohit Saraf  wrote:

> @Satya: I don't think what you have coded will work.. though i have not
> read the whole code.
>
> Don't you think a simple divide and conquer scheme would work...(almost
> like the mergesort)
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>
>
>
> On Sat, Jun 12, 2010 at 9:06 PM, Satya  wrote:
>
>> This problem seems to be finding the maxdiff in an array.
>>
>> int maxdiff ( int A[], int n ) {
>> // write your code here
>> int max_diff = A[1] - A[0];
>>   int min_element = A[0];
>>   int i;
>>   for(i = 1; i < n; i++)
>>   {
>> if(A[i] - min_element > max_diff)
>>   max_diff = A[i] - min_element;
>> if(A[i] < min_element)
>>  min_element = A[i];
>>   }
>>   if(max_diff < 0)
>> max_diff  = 0;
>>   return max_diff;
>> }
>>
>> .
>> Satya
>>
>>
>>
>> On Sat, Jun 12, 2010 at 3:18 PM, divya jain wrote:
>>
>>> i think we need to maintain an array of index as well such that while
>>> subtracting smallest element from largest element of sorted array we need to
>>> check if index of largest is greater than index of smallest. if no..then
>>> this is not the solution..
>>>
>>>
>>> On 12 June 2010 14:20, Modeling Expert wrote:
>>>
>>>> Let's say array A , 1 till n
>>>>
>>>> s_index = 1;  e_index = n ;
>>>> start  = &A[s_index] ;
>>>> end = &A[e_index];
>>>> result = 0;  //!  j - i
>>>> if ( *end > *start ){
>>>>result =  index(end) - index(start)  ( only of its greater than
>>>> previous value of result )
>>>>break ;
>>>> }else{
>>>> end-- ;  //! till you reach start
>>>> }
>>>>
>>>> now increment start , and repeat the process but only from A[n] till
>>>> A[++result] , going further
>>>> down is not required now.
>>>>
>>>> Average time < o(n^2)
>>>>
>>>> Worst case : let's say we have descending array of ints, theno(n^2)
>>>>
>>>> Please suggest improvements
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
>>>> On Jun 12, 12:14 am, amit  wrote:
>>>> > given an array A of n elements.
>>>> > for indexes j , i such that j>i
>>>> > maximize( j - i )
>>>> > such that A[j] - A [ i ]> 0 .
>>>> >
>>>> > Any Algorithm less than O(n^2) would do.
>>>>
>>>> --
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Re: [algogeeks] Mirroring Binary Tree Pattern Problem

[divya jain]

This C code will create a new mirror copy tree.
mynode *copy(mynode *root)
{
mynode *temp;
if(root==NULL)return(NULL);
temp = (mynode *) malloc(sizeof(mynode));
temp->value = root->value;
temp->left = copy(root->right);
temp->right = copy(root->left);
return(temp);
}

On 13 June 2010 17:07, BALARUKESH SIVARAMAN wrote:

> try this one..
> make a level order traversal and store the elements in array... on the
> other system reconstruct it using right element for the left and left
> element for the right...
>
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Re: [algogeeks] Re: bits

[divya jain]

thanks to all :)
@ souravsain sorry for the inconvenience


On 13 June 2010 20:01, Rohit Saraf  wrote:

> @Souravsain : Is there any serious problem in this. Anyone can just add a
> [C++] in the subject and uninterested people can make filters in gmail :)
>
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
>
> On Sun, Jun 13, 2010 at 6:35 PM, souravsain  wrote:
>
>> @divya
>>
>> Lets keep discussions in t his group limited to Algos and problems
>> neutral to any language.
>>
>> Request you to post these C++ / C language specific questions to those
>> language specific groups. This will not only help this group remain
>> confined to its core purpose but will help you get better insights to
>> ur problem by language specific geeks there. For C++ I would recommend
>> you to try the group
>> http://groups.google.co.in/group/comp.lang.c++.moderated/topics?hl=en.
>>
>> Regards,
>> Sourav
>>
>> On Jun 13, 2:29 pm, divya  wrote:
>> > tell the o/p of following with explanations
>> >
>> > 1. #include
>> > int main()
>> > {
>> >   struct value
>> > {
>> > int bit1:1;
>> > int bit3:4;
>> > int bit4:4;
>> >
>> > }bit;
>> >
>> > printf("%d\n",sizeof(bit));
>> > return 0;
>> >
>> > }
>> >
>> > 2.
>> > #include
>> > int main()
>> > {
>> > struct value
>> > {
>> > int bit1: 1;
>> > int bit3: 4;
>> > int bit4: 4;} bit={1,2,2};
>> >
>> > printf("%d %d %d\n",bit.bit1,bit.bit3,bit.bit4);
>> > return 0;
>> >
>> > }
>> >
>> > 3 can bit field be used in union??
>>
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Re: [algogeeks] file handing

[divya jain]

sorry i pasted wrong questn unser 2..

the real question is
which file will get closed through fclose()
#include
int main()
{
FILE *fp,*fs,*ft;
fp=fopen("a.c","r");
fs=fopen("b.c","r");
ft=fopen("c.c","r");
fclose(fp,fs,ft);
return 0;
}

3. yes it is feof..srry typed it wrong... nd fgets(str,80,fp) is perfectly
fine.. now the ans to this questn is that last line of the file will be
printed twice...( which i m unable to get why)...plzz explain...

@ souravsain plzz ignore this mail..srry for the inconvenience..

On 13 June 2010 17:37, jalaj jaiswal  wrote:

> in question 1... ch gets the value of EOF... so first kicit 44-a
> gokulpeth\0 nagpur will get printed and then the value of EOF..
>
>  question number 2 .. seems to me as nrml ...i think myfile.c only gets
> closed
>
> in question number 3..it shld be fgets(str,79,fp)
>
>
> On Sun, Jun 13, 2010 at 2:49 PM, divya  wrote:
>
>> 1. wat ll be the o/p. plz explain y?
>> // abc.c contains "kicit 44-a gokulpeth\0 nagpur"
>> #include
>>  #include
>>  int main()
>>  {
>>  unsigned char ch;
>>  FILE *fp;
>>  fp=fopen("abc.c","r");
>>  if(fp==NULL)
>>  {
>>  printf("unable to open the file \n");
>>  exit(1);
>>  }
>>  while((ch=getc(fp))!=EOF)
>>  printf("%c",ch);
>>  fclose(fp);
>>  printf("\n",ch);
>>  return 0;
>>  }
>>
>>  2.which file will get closed through fclose() in the following
>> program and why?
>> #include
>>  int main()
>>  {FILE *fp;
>>  char ch;
>>  int i=1;
>>  fp=fopen(myfile.c","r");
>>  while((ch=getc(fp)!=EOF))
>>  {
>>  if(ch=='\n')
>>  i++;
>>  }
>>
>>  fclose(fp);
>>  return 0;
>>  }
>>
>>  3.point out the error if any in following
>>
>> #include
>>  int main()
>>  {
>>  FILE *fp;
>>  char str[80];
>>  fp=fopen("trial","r");
>>  while(!eof(fp))
>>  {
>>  fgets(str,80,fp);
>>  puts(str);
>>  }
>>  fclose(fp);
>>  return 0;
>>  }
>>
>> --
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>
>
> --
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> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
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Re: [algogeeks] union- c

[divya jain]

wat i meant is
the ans of this questn is
10.00 0.00 3 1080263967
now my questn is y u.f_e is printing 0.00 and similarly y u.l_e is
giving this value...
On 13 June 2010 15:08, Rohit Saraf  wrote:

> If you are not able to print the long int and that's the prob, you can use
> %ld instead of %d
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>
>
> On Sun, Jun 13, 2010 at 1:49 PM, divya jain wrote:
>
>> its an o/p questn..
>> conversion wen ur variable is long..nd u r printing using %f...i dont know
>> how to perform conversion from float to int long nd vice versa..
>> pl help
>>
>> On 13 June 2010 12:12, Rohit Saraf  wrote:
>>
>>> what is this for...
>>> and which conversion are you talking abt?
>>>
>>> --
>>> Rohit Saraf
>>> Second Year Undergraduate,
>>> Dept. of Computer Science and Engineering
>>> IIT Bombay
>>> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>>>
>>>
>>> On Sat, Jun 12, 2010 at 11:20 PM, divya wrote:
>>>
>>>> #include 
>>>> main()
>>>> {
>>>>  union {
>>>>  long l_e;
>>>>  float f_e;
>>>>  } u;
>>>>
>>>>  long l_v;
>>>>  float f_v;
>>>>  l_v = u.l_e = 10;
>>>>  printf("%f ", (float)l_v);
>>>>  printf("%f ", u.f_e);
>>>>  f_v = u.f_e = 3.555;
>>>>  printf("%d ", (long)f_v);
>>>>  printf("%d ", u.l_e);
>>>> }
>>>> hw to do the conversion here..
>>>>
>>>> --
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>>>>
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Re: [algogeeks] c- pointers

[divya jain]

oh yes..
wen ll i stop making this stupid mistakes :(

On 13 June 2010 15:03, Rohit Saraf  wrote:

> @jalaj: exactly...
> so you(@divya) are right. Sharad's ans was right but logic wasn't.
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>
>
> On Sun, Jun 13, 2010 at 2:35 PM, jalaj jaiswal 
> wrote:
>
>> actually when you subtract two pointers ... its get divided by the size of
>> the variable type its point two...
>> for example.. if you do .. p+1... where let say p is 200 and points to an
>> int type variable then p+1 is 202...(assuming int is of size 2)
>> so (p+1)-p..i.e 202-200 is 1 and not 2
>>
>>
>>
>>
>>
>>
>> On Sun, Jun 13, 2010 at 1:46 PM, divya jain wrote:
>>
>>> bt the ans that sharad gave is ryt..
>>> acc to me 1st row n 1st col of o/p shd b 2 (if size of int is 2) bt it is
>>> 1...
>>>
>>>
>>> On 13 June 2010 12:10, Rohit Saraf  wrote:
>>>
>>>> @divya: u r rite.. that * should not be there
>>>>
>>>> --
>>>> Rohit Saraf
>>>> Second Year Undergraduate,
>>>> Dept. of Computer Science and Engineering
>>>> IIT Bombay
>>>> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>>>>
>>>>
>>>> On Sun, Jun 13, 2010 at 11:07 AM, divya jain 
>>>> wrote:
>>>>
>>>>> ptr is a pointer naaa...then why ptr-p=*(&(arr+1)-&arr)  ???
>>>>> why not &(arr+1)-&arr ??
>>>>> i knw m wrong somewhr...plz correct me
>>>>>
>>>>>
>>>>> On 13 June 2010 07:57, Mahesh_JNU  wrote:
>>>>>
>>>>>> agreed .
>>>>>>
>>>>>>
>>>>>> On Sun, Jun 13, 2010 at 7:48 AM, sharad kumar <
>>>>>> aryansmit3...@gmail.com> wrote:
>>>>>>
>>>>>>> 111
>>>>>>> 222
>>>>>>> 333
>>>>>>> 344
>>>>>>> ptr++ ->u do posst increment
>>>>>>> hence it goes to 1
>>>>>>> ptr-p=*(&(arr+1)-&arr)=1
>>>>>>> llrly for other cases
>>>>>>> when u do *ptr++ due to operator precedence ptr++ is done and then
>>>>>>> dereferenced.
>>>>>>> hence u get 222
>>>>>>> next *++ptr
>>>>>>> the ptr is incremented after dereferencing hence u get 333
>>>>>>> next ++*ptr here the value t ptr s incrementas it is treated
>>>>>>> as++(*ptr) hence u get 3 but others refer to location hence 44
>>>>>>>
>>>>>>>
>>>>>>> On Sat, Jun 12, 2010 at 9:21 PM, divya wrote:
>>>>>>>
>>>>>>>> #include
>>>>>>>> int main()
>>>>>>>> {
>>>>>>>> static int arr[]={0,1,2,3,4};
>>>>>>>> int *p[]={arr,arr+1,arr+2,arr+3,arr+4};
>>>>>>>> int **ptr=p;
>>>>>>>> ptr++;
>>>>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>>>>> *ptr++;
>>>>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>>>>> *++ptr;
>>>>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>>>>> ++*ptr;
>>>>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>>>>> return 0;
>>>>>>>> }
>>>>>>>> wat shd b the o/p n why...
>>>>>>>>
>>>>>>>> --
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>>>>>>>> http://groups.google.c

Re: [algogeeks] union- c

[divya jain]

its an o/p questn..
conversion wen ur variable is long..nd u r printing using %f...i dont know
how to perform conversion from float to int long nd vice versa..
pl help

On 13 June 2010 12:12, Rohit Saraf  wrote:

> what is this for...
> and which conversion are you talking abt?
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
>
> On Sat, Jun 12, 2010 at 11:20 PM, divya  wrote:
>
>> #include 
>> main()
>> {
>>  union {
>>  long l_e;
>>  float f_e;
>>  } u;
>>
>>  long l_v;
>>  float f_v;
>>  l_v = u.l_e = 10;
>>  printf("%f ", (float)l_v);
>>  printf("%f ", u.f_e);
>>  f_v = u.f_e = 3.555;
>>  printf("%d ", (long)f_v);
>>  printf("%d ", u.l_e);
>> }
>> hw to do the conversion here..
>>
>> --
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Re: [algogeeks] c array

[divya jain]

i use tc

On 13 June 2010 13:11, ram  wrote:

> @rohit bro
>
> http://www.mingw.org/
>
> *MinGW*, a contraction of "Minimalist GNU for Windows", is a port of the
> GNU Compiler Collection (GCC), and GNU Binutils, for use in the development
> of native Microsoft Windows applications.
>
>
>
>
>
> *From:* algogeeks@googlegroups.com [mailto:algoge...@googlegroups.com] *On
> Behalf Of *Rohit Saraf
> *Sent:* 13 June 2010 08:19
>
> *To:* algogeeks@googlegroups.com
> *Subject:* Re: [algogeeks] c array
>
>
>
> @ram : i guess you have used some longer string and not "strings"
>
>
>
> btw..  what is Mingw ?
>
> gcc/g++ is not mingw, i guess
>
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
> On Sun, Jun 13, 2010 at 8:13 AM, ram  wrote:
>
> D:\code\samplecode\main.cpp|5|error: initializer-string for array of chars
> is too long|
>
>
>
> I get this error on gcc (Mingw) .
>
>
>
> Though the array indexing starts from 0.
>
> The length specified in char str[7] is always straightforward . in this
> case char str[7]  . the length of str is seven not eight ;hence the error
>
> --
>
> ram
>
>
>
> *From:* algogeeks@googlegroups.com [mailto:algoge...@googlegroups.com] *On
> Behalf Of *sharad kumar
> *Sent:* 13 June 2010 07:59
> *To:* algogeeks@googlegroups.com
> *Subject:* Re: [algogeeks] c array
>
>
>
> hey array indexing starts from 0 rite??
> then y shld u get overflow in first place..
> s t  r  i n g s \0
> 0 1 2 3 4 5 6 7
>
> On Sat, Jun 12, 2010 at 9:14 PM, divya  wrote:
>
> #include
> int main()
> {
>
> char str[7]="strings";
> printf("%s\n",str);
> return 0;
> }
>
> here i m nt getting overflow error whereas if i write stringss instead
> of strings then there is overflow error.. isnt null stored after s in
> strings nd 1st case shd also give overflow???
>
> --
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Re: [algogeeks] c- pointers

[divya jain]

bt the ans that sharad gave is ryt..
acc to me 1st row n 1st col of o/p shd b 2 (if size of int is 2) bt it is
1...

On 13 June 2010 12:10, Rohit Saraf  wrote:

> @divya: u r rite.. that * should not be there
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>
>
> On Sun, Jun 13, 2010 at 11:07 AM, divya jain wrote:
>
>> ptr is a pointer naaa...then why ptr-p=*(&(arr+1)-&arr)  ???
>> why not &(arr+1)-&arr ??
>> i knw m wrong somewhr...plz correct me
>>
>>
>> On 13 June 2010 07:57, Mahesh_JNU  wrote:
>>
>>> agreed .
>>>
>>>
>>> On Sun, Jun 13, 2010 at 7:48 AM, sharad kumar 
>>> wrote:
>>>
>>>> 111
>>>> 222
>>>> 333
>>>> 344
>>>> ptr++ ->u do posst increment
>>>> hence it goes to 1
>>>> ptr-p=*(&(arr+1)-&arr)=1
>>>> llrly for other cases
>>>> when u do *ptr++ due to operator precedence ptr++ is done and then
>>>> dereferenced.
>>>> hence u get 222
>>>> next *++ptr
>>>> the ptr is incremented after dereferencing hence u get 333
>>>> next ++*ptr here the value t ptr s incrementas it is treated as++(*ptr)
>>>> hence u get 3 but others refer to location hence 44
>>>>
>>>>
>>>> On Sat, Jun 12, 2010 at 9:21 PM, divya wrote:
>>>>
>>>>> #include
>>>>> int main()
>>>>> {
>>>>> static int arr[]={0,1,2,3,4};
>>>>> int *p[]={arr,arr+1,arr+2,arr+3,arr+4};
>>>>> int **ptr=p;
>>>>> ptr++;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> *ptr++;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> *++ptr;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> ++*ptr;
>>>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>>>> return 0;
>>>>> }
>>>>> wat shd b the o/p n why...
>>>>>
>>>>> --
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>>>>>
>>>>
>>>>
>>>> --
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>>>
>>>
>>>
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Re: [algogeeks] Re: identify the recurring part for a given decimal no

[divya jain]

thanks anurag :)

On 12 June 2010 20:07, Anurag Sharma  wrote:

> Since we are given numerator 'n' and denominator 'd' separately already.
> and considering n and d as integers and d!=0 we can safely assume n/d as
> either a terminating fraction or a non terminating but recurring fraction,
> in which case we have to find the recurring digits of the fraction.
>
> Now what I suggested was almost same as Ravi's approach.
> take a Set 'S' keeping tuples (R,Q) where R is the current remainder and Q
> is the factor such that d*Q is subtracted from the number to get R.
> In other words. if at an intermediate step of division we have 'a' as the
> divident left then Q=floor(a/d) and R=a%d
>
> Keep dividing 'n' by 'd' like it is done manually. After every division
> check-
> 1. If the current remainder is not present in 'S' then add current
> remainder 'R' and corresponding quotient 'Q' in the set
> 2. If R is found in the set S, then all the following entries in the set
> until end will constitute the recurring digits.
> taking Ravi's example:-
>
> Example:
>   7) 9 (1.*285714*28S=[]
>7
>--
> 20   S=[(2,2)]
>  14
>  ---
>60S=[(2,2), (6,8)]
> 56
>  ---
>   40 S=[(2,2), (6,8),
> (4,5)]
>   35
>   ---
>  50  S=[(2,2), (6,8),
> (4,5), (5,7)]
>   49
>---
>  10  S=[(2,2), (6,8),
> (4,5), (5,7), (1,1)]
> 7
>  
>  30  S=[(2,2), (6,8),
> (4,5), (5,7), (1,1), (3,4)]
>   28   ^
>      |
>   20 2 is found in S here,
> so recurring digits are "285714"
>    14
>
>   
>60
> 56
>  repeats
>
>
> hope its clear
>
>
> Anurag Sharma
>
>
> On Sat, Jun 12, 2010 at 4:02 PM, divya jain wrote:
>
>> @anurag
>>
>> i dint get ur approach..which numerator n denominator u r talking
>> about..plz explain.. thanks in advance
>>
>> On 11 June 2010 08:57, Anurag Sharma  wrote:
>>
>>> Please note that the fractional repeating part is recurring. and so that
>>> 4th temporary variable assignment will be this way->
>>> temp=x*1 - x= 233456.34563456...  - 23.34563456 = 233433.0  (
>>> mark the fractional part is 0 now since the infinitely repeating 3456...
>>> will get cancelled)
>>> In this  case you can say that 4 places are repeating. But yes its
>>> according to the maths and in any programming language whenever you divide
>>> the numerator and denominator you wont get this infinitely recurring decimal
>>> places.
>>>
>>> @divya, also your approach wont work if the recurring fractional digits
>>> start after few places from the decimal like in the case of
>>> 23.123345634563456  (note here after the decimal place 123 is not
>>> repeating while 3456.. after this 123 is repeating.)
>>>
>>> What I suggest in this case is keep dividing the numerator by denominator
>>> and at every step keep inserting the tupple (remainder, quotient) of that
>>> division step in a set. and before inserting in the set check whether it
>>> already exists. If yes then the all the quotients following from that point
>>> (including the point) will be recurring.
>>>
>>> Regards,
>>>
>>> Anurag Sharma
>>>
>>>
>>>
>>> On Thu, Jun 10, 2010 at 8:25 AM, Veer Sharma 
>>> wrote:
>>>
>>>> Seems it wont work...
>>>> x=23.34563456
>>>>
>>>> temp = x*100 -x = 233.4563456 - 23.34563456 = 210.11071104
>>>> temp = x*100 -x = 2334.563456 - 23.34563456 = 2311.21782144
>>>> temp = x*1000 -x =  23345.63456 - 23.34563456 = 23322.28892544
>>>> temp = x*1 -x =  233456.3456 - 23.34563456 = 233432.6544
>>>> temp = x*10 -x = 2334563.456 - 23.34563456 = 2334540.11036544
>>>>
>>>> ...
>

Re: [algogeeks] c- pointers

[divya jain]

ptr is a pointer naaa...then why ptr-p=*(&(arr+1)-&arr)  ???
why not &(arr+1)-&arr ??
i knw m wrong somewhr...plz correct me

On 13 June 2010 07:57, Mahesh_JNU  wrote:

> agreed .
>
>
> On Sun, Jun 13, 2010 at 7:48 AM, sharad kumar wrote:
>
>> 111
>> 222
>> 333
>> 344
>> ptr++ ->u do posst increment
>> hence it goes to 1
>> ptr-p=*(&(arr+1)-&arr)=1
>> llrly for other cases
>> when u do *ptr++ due to operator precedence ptr++ is done and then
>> dereferenced.
>> hence u get 222
>> next *++ptr
>> the ptr is incremented after dereferencing hence u get 333
>> next ++*ptr here the value t ptr s incrementas it is treated as++(*ptr)
>> hence u get 3 but others refer to location hence 44
>>
>>
>> On Sat, Jun 12, 2010 at 9:21 PM, divya  wrote:
>>
>>> #include
>>> int main()
>>> {
>>> static int arr[]={0,1,2,3,4};
>>> int *p[]={arr,arr+1,arr+2,arr+3,arr+4};
>>> int **ptr=p;
>>> ptr++;
>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>> *ptr++;
>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>> *++ptr;
>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>> ++*ptr;
>>> printf("%d %d %d\n",ptr-p,*ptr-arr,**ptr);
>>> return 0;
>>> }
>>> wat shd b the o/p n why...
>>>
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>>>
>>
>>
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>
>
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Re: [algogeeks] c array

[divya jain]

hmm..the prob is here wid my compiler...!!
anyways thanks to all

On 13 June 2010 10:28, Rohit Saraf  wrote:

> oh.. yes.. my mistake
> (strings\0).length=8 :P
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
>
> On Sun, Jun 13, 2010 at 10:24 AM, Rahul Kushwaha  > wrote:
>
>> #include
>> int main()
>> {
>> char str[7]="strings";
>> printf("%s\n",str);
>> return 0;
>> }
>>
>>
>> it is showing error on code block and dev cpp also...
>> this  is an error no doubt.
>> also mentioned in denis m ritchie
>>
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Re: [algogeeks] output

[divya jain]

sorry for the silly question i got rhe point..

@ rohit
compiler is doing rite..read mahesh's explanatn

On 13 June 2010 08:27, Rohit Saraf  wrote:

> This is very bad. Change your compiler if it compiles this stuff :)
>
> btw.. which compiler is it?
>
> Output for me :
> ro...@rohit-laptop:~/dump$ gcc c.c
> c.c: In function ‘main’:
> c.c:14: error: incompatible types when assigning to type ‘char[20]’ from
> type ‘char *’
> c.c:15: error: incompatible types when assigning to type ‘char[20]’ from
> type ‘char *’
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
>
>
> On Sun, Jun 13, 2010 at 8:13 AM, Mahesh_JNU wrote:
>
>> Well
>>
>> As we know for copying the string we can can copy it as a simple variable
>> as in case of address copying.
>> when u r doing names[3] = names[4] , it means u r trying to copy it
>> directly
>> bt in the case of  char *names[] , as it is the array of pointers so u can
>> copy the address from one pointer to another pointer
>>
>> Thanks
>>
>>
>> On Sat, Jun 12, 2010 at 9:12 PM, divya  wrote:
>>
>>> #include
>>>
>>> int main()
>>> { char names[][20]={
>>> "roshni",
>>> "manish",
>>> "sona",
>>> "baiju",
>>> "ritu"
>>> };
>>> int i;
>>> char *t;
>>> t=names[3];
>>> names[3]=names[4];
>>> names[4]=t;
>>> for(i=0;i<=4;i++)
>>> printf("%s",names[i]);
>>> printf("\n");
>>> return 0;
>>> }
>>>
>>> here i get l value required as error and if i replace char names[][2]
>>> with char *names[].. then there is no error nd the names[3] n names[4]
>>> interchange
>>> pl explain why???
>>>
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>>
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Re: [algogeeks] Re: c output

[divya jain]

thanks to all for explanantions :)

On 12 June 2010 15:43, divya jain  wrote:

> one of my frnd askd me this question...
>
>
> On 11 June 2010 21:34, Raj N  wrote:
>
>> @kirubakaran: How can it be 1,1 ? No of characters read in a is 5+ 1 for
>> '\n' so its 6 and for the next one 1+1=2
>>
>>
>> On Fri, Jun 11, 2010 at 9:09 AM, kirubakaran 
>> wrote:
>>
>>> Output will be
>>>
>>> 1,1
>>> bcoz printf returns number of characters or integers printed
>>>
>>> On Jun 11, 12:26 am, divya  wrote:
>>> > #include 
>>> > main()
>>> > {
>>> >  int a = 1;
>>> >  char b='c';
>>> >  int i,j;
>>> >
>>> >  printf("%d,%d",printf("%d\n",a),printf("%c\n",b));
>>> >
>>> > wat shd b the o/p of this..plzz explain y?
>>>
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Re: [algogeeks] Re: identify the recurring part for a given decimal no

[divya jain]

@anurag

i dint get ur approach..which numerator n denominator u r talking about..plz
explain.. thanks in advance

On 11 June 2010 08:57, Anurag Sharma  wrote:

> Please note that the fractional repeating part is recurring. and so that
> 4th temporary variable assignment will be this way->
> temp=x*1 - x= 233456.34563456...  - 23.34563456 = 233433.0  ( mark
> the fractional part is 0 now since the infinitely repeating 3456... will get
> cancelled)
> In this  case you can say that 4 places are repeating. But yes its
> according to the maths and in any programming language whenever you divide
> the numerator and denominator you wont get this infinitely recurring decimal
> places.
>
> @divya, also your approach wont work if the recurring fractional digits
> start after few places from the decimal like in the case of
> 23.123345634563456  (note here after the decimal place 123 is not
> repeating while 3456.. after this 123 is repeating.)
>
> What I suggest in this case is keep dividing the numerator by denominator
> and at every step keep inserting the tupple (remainder, quotient) of that
> division step in a set. and before inserting in the set check whether it
> already exists. If yes then the all the quotients following from that point
> (including the point) will be recurring.
>
> Regards,
>
> Anurag Sharma
>
>
>
> On Thu, Jun 10, 2010 at 8:25 AM, Veer Sharma wrote:
>
>> Seems it wont work...
>> x=23.34563456
>>
>> temp = x*100 -x = 233.4563456 - 23.34563456 = 210.11071104
>> temp = x*100 -x = 2334.563456 - 23.34563456 = 2311.21782144
>> temp = x*1000 -x =  23345.63456 - 23.34563456 = 23322.28892544
>> temp = x*1 -x =  233456.3456 - 23.34563456 = 233432.6544
>> temp = x*10 -x = 2334563.456 - 23.34563456 = 2334540.11036544
>>
>> ...
>>
>> On Jun 9, 11:24 pm, Anurag Sharma  wrote:
>> > multiply the original number x=23.34563456
>> >
>> > Anurag Sharma
>> >
>> > On Wed, Jun 9, 2010 at 10:36 PM, Veer Sharma > >wrote:
>> >
>> >
>> >
>> > > One question:
>> >
>> > > No x = 23.34563456
>> > > temp = x X 10 = 233.4563456
>> > > temp = temp - x = 210.11071104
>> > > decimal part zero? No.
>> > > Now multiply the no. with 100. Which no? original x (= 23.34563456) or
>> > > new no. temp (=210.11071104)?
>> >
>> > > On Jun 9, 8:12 pm, divya jain  wrote:
>> > > > multiply the no. with 10 nd store in temp. now subtract no from
>> temp.
>> > > check
>> > > > if the decimal part is zero if yes.  then 1st digit after decimal is
>> > > > recurring. if no. multiply the no with 100 and repeat . if this time
>> > > decimal
>> > > > part is zero then 2 digits after decimal r recurring nd so on..
>> >
>> > > > On 8 June 2010 21:45, Veer Sharma 
>> wrote:
>> >
>> > > > > You have a Numerator and Denominator. After division you might get
>> a
>> > > > > recurring decimal points float as the answer.
>> >
>> > > > > Problem is: You need to identify the recurring part for a given
>> > > > > decimal no?
>> > > > > For example 23.34563456 ...
>> > > > > return 3456
>> >
>> > > > > --
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Re: [algogeeks] Re: c output

[divya jain]

one of my frnd askd me this question...

On 11 June 2010 21:34, Raj N  wrote:

> @kirubakaran: How can it be 1,1 ? No of characters read in a is 5+ 1 for
> '\n' so its 6 and for the next one 1+1=2
>
>
> On Fri, Jun 11, 2010 at 9:09 AM, kirubakaran wrote:
>
>> Output will be
>>
>> 1,1
>> bcoz printf returns number of characters or integers printed
>>
>> On Jun 11, 12:26 am, divya  wrote:
>> > #include 
>> > main()
>> > {
>> >  int a = 1;
>> >  char b='c';
>> >  int i,j;
>> >
>> >  printf("%d,%d",printf("%d\n",a),printf("%c\n",b));
>> >
>> > wat shd b the o/p of this..plzz explain y?
>>
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Re: [algogeeks] Re: Array Problem

[divya jain]

i think we need to maintain an array of index as well such that while
subtracting smallest element from largest element of sorted array we need to
check if index of largest is greater than index of smallest. if no..then
this is not the solution..

On 12 June 2010 14:20, Modeling Expert  wrote:

> Let's say array A , 1 till n
>
> s_index = 1;  e_index = n ;
> start  = &A[s_index] ;
> end = &A[e_index];
> result = 0;  //!  j - i
> if ( *end > *start ){
>result =  index(end) - index(start)  ( only of its greater than
> previous value of result )
>break ;
> }else{
> end-- ;  //! till you reach start
> }
>
> now increment start , and repeat the process but only from A[n] till
> A[++result] , going further
> down is not required now.
>
> Average time < o(n^2)
>
> Worst case : let's say we have descending array of ints, theno(n^2)
>
> Please suggest improvements
>
>
>
>
>
>
>
>
>
>
> On Jun 12, 12:14 am, amit  wrote:
> > given an array A of n elements.
> > for indexes j , i such that j>i
> > maximize( j - i )
> > such that A[j] - A [ i ]> 0 .
> >
> > Any Algorithm less than O(n^2) would do.
>
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Re: [algogeeks] Re: isomorphic trees

[divya jain]

my solution
*isomorphic* (struct node *head1,struct node *head2)
{
if (head1!=NULL && head2!=NULL)
{
if (head1->data == head2->data)
{
if (head1->left->data == head2->left->data)
{
return (isomorphic(head1>left,head2->left ) &&
isomorphic(head1>right,head2->right));
}
else if (head->left->data == head2->right->data)
{
return (isomorphic**(head1>left,head2->right)
&&(isomorphichead1>right,head2->left));
}
else
return (false);
}
}
else if (head1 == NULL && head2 == NULL)
return (true);
else
return (false);
}

On 10 June 2010 00:00, souravsain  wrote:

>
> As per @Algoose's explanation, this can be found using the algorithm
> to comapre if 2 binary trees are equal (quite often asked and found in
> net). In this we generally go recursive and say
> T1 is equal to T2
> if T1=T2
> and same holds for T1-Left, T2-Left (recursive call on left tree)
> and same holds for T1-Right, T2-Right (recursive call on right tree).
>
> We can use the same and change the calls as
> if T1=T2
> and same holds for T1-Left, T2-Right (recursive call on left tree)
> and same holds for T1-Right, T2-Left (recursive call on right tree).
>
>
> Sain
>
> On Jun 9, 10:45 pm, saurabh gupta  wrote:
> > is-isomorphic(t1, t2) {
> >  t1ld = t1->left->data
> >  t2ld = t2->left->data
> > //.
> >
> > //base case for null or replace by sentinels and check
> > if( t1ld == t2ld && t1rd == t2rd)
> >return is-isomorphic(t1ld, t2ld) && return is-isomorphic(t1rd,
> t2rd)
> > if (t1ld == t2rd && t1rd == t2ld)
> >return is-isomorphic(t1ld, t2rd) && return is-isomorphic(t1rd,
> t2ld)
> > return false;
> >
> >
> >
> >
> >
> > }
> > On Wed, Jun 9, 2010 at 8:29 PM, divya jain 
> wrote:
> > > @vadivel and anand
> >
> > > { 1,2,3 } is *isomorphic* to { 1,3,2 }
> > > { 1,2,3,4,5,6,7 } is *isomorphic* to { 1,3,2,7,6,4,5 }
> > > { 1,2,3,4,5,6,7 } is NOT *isomorphic* to { 1,3,2,4,5,6,7 }
> >
> > > so its nt necessary that right and left will interchange. it may or may
> > > not. if all right and left are interchanged then it is mirror tree. i
> think
> > > ur code is for mirror tree and not isomorphic tree..
> >
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> > Says he feels all alone in a threatening world where what lies ahead is
> > vague and uncertain. Doctor says "Treatment is simple. Great clown
> > Pagliacci is in town tonight. Go and see him. That should pick you up."
> Man
> > bursts into tears. Says "But, doctor...I am Pagliacci."- Hide quoted text
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> > - Show quoted text -
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Re: [algogeeks] isomorphic trees

[divya jain]

@vadivel and anand

{ 1,2,3 } is *isomorphic* to { 1,3,2 }
{ 1,2,3,4,5,6,7 } is *isomorphic* to { 1,3,2,7,6,4,5 }
{ 1,2,3,4,5,6,7 } is NOT *isomorphic* to { 1,3,2,4,5,6,7 }

so its nt necessary that right and left will interchange. it may or may not.
if all right and left are interchanged then it is mirror tree. i think ur
code is for mirror tree and not isomorphic tree..

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Re: [algogeeks] identify the recurring part for a given decimal no

[divya jain]

multiply the no. with 10 nd store in temp. now subtract no from temp. check
if the decimal part is zero if yes.  then 1st digit after decimal is
recurring. if no. multiply the no with 100 and repeat . if this time decimal
part is zero then 2 digits after decimal r recurring nd so on..

On 8 June 2010 21:45, Veer Sharma  wrote:

> You have a Numerator and Denominator. After division you might get a
> recurring decimal points float as the answer.
>
> Problem is: You need to identify the recurring part for a given
> decimal no?
> For example 23.34563456 ...
> return 3456
>
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Re: [algogeeks] min no of policemen

[divya jain]

edge is the link connecting two nodes of a tree

On 8 June 2010 11:23, Anurag Sharma  wrote:

> Can you expain  "edge of the tree". I could not get that.
>
> Anurag Sharma
>
>
>
> On Tue, Jun 8, 2010 at 5:53 AM, sharad kumar wrote:
>
>> for placing police man we can use bellman ford bfs.or even make use of
>> topological sort.
>>
>>
>> On Mon, Jun 7, 2010 at 9:59 PM, divya  wrote:
>>
>>> consider a tree. policemen is to be placed such that for each edge of
>>> tree, there is a policeman on atleast one side of each edge. tell the
>>> min no. of policemen and their locatn in time O(n)
>>>
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>>
>>
>> --
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>>
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Re: [algogeeks] Single ordered list

[divya jain]

merging as in merge sort.

On 8 June 2010 19:59, Raj N  wrote:

> What is the most efficient way of combining 2 separate ordered list
> into a single ordered list ?
>
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Re: [algogeeks] Puzzle:

[divya jain]

yes even i dont think its possible as there is no n-1th state..ie there is
no state from whr u can come to 2 8 5..so plz check

On 7 June 2010 20:23, mohit ranjan  wrote:

> is it possible ?
>
> if we say nth state is [2, 8, 5]
>
> I could not find possible (n-1)th state
>
>
> Mohit Ranjan
>
>
>
> On Mon, Jun 7, 2010 at 2:02 PM, sharad  wrote:
>
>> : Three containers are of 15,10 and 6 ltrs capacity. Initially its in
>> configuration (15,0,0). Make it to configuration (2,8,5)
>>
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Re: [algogeeks] Pointer to a constant

[divya jain]

i think both statements shd give error. as u r trying to change int to const
int in 2 and const int to int in 1..

On 7 June 2010 19:59, mohit ranjan  wrote:

> @Raj,
>
> no they are not same
>
>
> case 1: i is const
> case 2: ptr is const
>
> and whatever is const cann't be modified
>
> Mohit Ranjan
>
>
> On Mon, Jun 7, 2010 at 3:01 PM, Raj N  wrote:
>
>> Can someone tell me the difference between
>> 1) const int i=5; 2)  int i=5;
>>   int *ptr=&i;  const int
>> *ptr=&i;
>>
>> In the first case i can be modified via ptr i.e *ptr++ is valid. In
>> the second case *ptr++ is illegal. Why is that so? Aren't they same?
>>
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Re: [algogeeks] Number of sequences of n binary digits that don't contain two 1's in a row

[divya jain]

how do u come to this formula T(n)=fib(n+2).. plz explain

On 7 June 2010 21:19, saurabh gupta  wrote:

> it might be referring to no of sequences (say T(n) ) with no consecutive
> 1's
> for n = 3, ans would be 5 viz.
> 000, 001, 010, 100, 101
>
> T(n) =  fib(n+2)
> where fib = Fibonacci series
> which is interesting.
>
>
> On Sat, Jun 5, 2010 at 11:40 AM, Raj N  wrote:
>
>> @sharad: What about 101 even it doesn't have two 1's in a row
>>
>>
>> On Sat, Jun 5, 2010 at 8:59 AM, sharad kumar wrote:
>>
>>> @rajn.can it be subsequence doesnt have one's too.hence 000,001,010,100
>>> is required answer.
>>>
>>>
>>> On Sat, Jun 5, 2010 at 12:13 AM, Raj N  wrote:
>>>
 Hi,
 I came across this question to find the number of sequences of n
 binary digits that don't contain 2 1's in a row. I wanted to know what
 exactly this means. Is it like if n=3 then compute all binary numbers
 having 3 digits which don't have consecutive 1's 110, 011, 111 ??
 If not help me understanding it.
 Thanks!!

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>>>
>>>
>>> --
>>> yezhu malai vaasa venkataramana Govinda Govinda
>>>
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>
>
>
> --
> Man goes to doctor. Says he's depressed. Says life seems harsh and cruel.
> Says he feels all alone in a threatening world where what lies ahead is
> vague and uncertain. Doctor says "Treatment is simple. Great clown
> Pagliacci is in town tonight. Go and see him. That should pick you up." Man
> bursts into tears. Says "But, doctor...I am Pagliacci."
>
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Re: [algogeeks] min no of policemen

[divya jain]

@sharad..

sorry bt i dint get how to use bellman ford or topological sort here...
can u plz explain...

On 8 June 2010 05:53, sharad kumar  wrote:

> for placing police man we can use bellman ford bfs.or even make use of
> topological sort.
>
>
> On Mon, Jun 7, 2010 at 9:59 PM, divya  wrote:
>
>> consider a tree. policemen is to be placed such that for each edge of
>> tree, there is a policeman on atleast one side of each edge. tell the
>> min no. of policemen and their locatn in time O(n)
>>
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Re: [algogeeks] matix flipping

[divya jain]

let a[n][n] be the input array nd b[][] be the output

for(i=0;i wrote:

> write a c routine to flip an nXn matrix about its non major diagnol
>
>
> 3   4   5
> 6   7   9
> 1   2   8
>
> Transpose is:
> 8   9   5
> 2   7   4
> 1   6   3
>
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Re: [algogeeks] Re: array question

[divya jain]

output willl be 12 12 5 6 6

On 6 June 2010 18:27, souravsain  wrote:

> @divya: Does your problem require the output to be sorted also? What
> will be the output required if inout is 12,5,6,12,6? Will it be
> 12,12,6,6,5 or 12,12,5,6,6,?
>
> Sain
>
> On Jun 6, 12:01 am, divya  wrote:
> > Given an array with some repeating numbers. Like 12,6,5,12,6
> >
> > output: 12,12,6,6,5
> > 12 shud come before 6 since it is earlier in list. So cant use a
> > dictionary
>
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Re: [algogeeks] array question

[divya jain]

@sharad
while storing each element in hash by your approach u ll check if its
already there in hash. so the complexity here will be O(n2). correct me if i
m wrong. isnt there ny better algo..?

On 6 June 2010 06:28, sharad kumar  wrote:

> @dhivya:keep storing the first occurance element index in hash map and then
> start insertin eleement based on index position
>
>
> On Sun, Jun 6, 2010 at 12:31 AM, divya  wrote:
>
>> Given an array with some repeating numbers. Like 12,6,5,12,6
>>
>> output: 12,12,6,6,5
>> 12 shud come before 6 since it is earlier in list. So cant use a
>> dictionary.
>>
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Re: [algogeeks] o/p

[divya jain]

turbo c++ and my o/p is logical as well

On 5 June 2010 17:58, sharad kumar  wrote:

> @divya,which compiler u r using,i m getting this o/p on gcc compiler
>
> @mohit:loop stops at 4,294,967,295 in gcc
>
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[algogeeks] dynamic vs greedy aproach

[divya jain]

how can we make out whether to apply greedy approach or dynamic programming
to a problem??

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Re: [algogeeks] Largest even length palindrome in a string!!

[divya jain]

1. first find two consecutive letters which are same.
2 point ptr1 to left character and ptr2 to right one
3. now increment ptr2 and decrement ptr1. compare if they are pointing to
same characters. if yes repeat this step
4. if no then store in length ptr2-ptr1-2
5. go to step 1 untill all consecutive same letters have been scanned. if
the new length found is greater then previous one. then discard previous one
and store in length new length.
6. return the substring with largest length


On 5 June 2010 15:04, Satya  wrote:

> Hi,
>
> How to find largest palindrome, which is even in length in a string.
> Complexity should be "lessthan" O(n^2).
>
> Ex;- abacbbcababac - Given string.
> 'abacbbcaba' - is the largest even length palindrome.
>
> Thanks,
> Satya
>
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Re: [algogeeks] o/p

[divya jain]

output on my compiler is 165535
as i=0 initialy ++i is true and so for loop condition is true and printf is
executed. when i becomes 65535 then ++i is equal to 0 and so for loop
condition becomes false.

On 5 June 2010 13:44, sharad  wrote:

> #include
> int main()
> {
> short int i=0;
> for(i<=5&&i>=-1;++i;i>0)
> printf("%u\n",i);
> return 0;
> }
>
> o/p coming is 1...4,294,967,295
> short int is of 2 bytes
> can any one plzz explain the o/p
>
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Re: [algogeeks] Quick short stack space reduction

[divya jain]

if u sort the larger one first one then smaller one will be on stack for a
long time on the other hand if u sort smaller one first then larger one will
be in stack for long time. so more space is wasted is 2nd case so we shd
sort larger first. correct me if i am wrong plzzz

On 4 June 2010 18:08, Rohit Saraf  wrote:

> if you "short" the larger one first, then the smaller one will be in the
> stack for a long time. Which is wasting stack space.
> Now stop "shorting" and start sorting ! :)
>
> --
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> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
>
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Re: [algogeeks] Re: partion of array

[divya jain]

oh...okay...
good example...

On 3 June 2010 13:23, Rohit Saraf  wrote:

> @divya: but still haven't you seen Jagadish's example?  It is a
> counterexample to your greedy approach.
>
> --
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Re: [algogeeks] Removing extra parentheses in an infix string

[divya jain]

1.calculte the postfix of given expression.
2.now remove a particular parenthesis from expression and check if the
postfix of this expression is equal to the postfix of original expression.
if yes then the parenthesis we have removed were extra. if no then the
parenthesis were not exta.
3 now remove other parenthesis as step 2 and repeat till u have done this
for all parenthesis

On 1 June 2010 20:12, Raj N  wrote:

> How to remove extra parentheses in an infix string. For example if it
> is A+(B*C) parentheses for * is not required as it has higher
> precedence. Can someone suggest a good routine for this?
>
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Re: [algogeeks] Re: Can you solve this?

[divya jain]

same question as wat i asked in partioning of array such that the diff is
min.

On 31 May 2010 22:07, Senthilnathan Maadasamy <
senthilnathan.maadas...@gmail.com> wrote:

> Same question with interesting answers in stackoverflow :
> http://stackoverflow.com/questions/890171/algorithm-to-divide-a-list-of-numbers-into-2-equal-sum-lists
>
>
>
> On Mon, May 31, 2010 at 5:54 PM, Anurag Sharma wrote:
>
>> Well, in that case, then just forget the "balancing the number of elements
>> in the 2 teams" portion of my solution above :)
>>
>>
>> Anurag Sharma
>> http://anuragsharma-sun.blogspot.com/
>>
>>
>> On Mon, May 31, 2010 at 10:38 AM, Nik_nitdgp 
>> wrote:
>>
>>> This problem is like 2 processor job scheduling problem ,We may get an
>>> optimal solution for different instances using different algorithm
>>> apart from brute force.Whereas Brute force covers all possible subsets
>>> but may take years to complete if N is large.
>>>
>>> above algo fails in the following example.
>>>
>>> Eg. 2 2 2 3 3
>>>
>>> above algo gives:
>>> T1: 2 2 3 =7
>>> T2: 2 3 =5
>>>
>>> But closest distribution is
>>> T1=2 2 2=6
>>> T2 3 3=6
>>>
>>> On May 31, 9:30 am, W Karas  wrote:
>>> > Is this the same problem as:
>>> >
>>> > http://groups.google.com/group/algogeeks/browse_frm/thread/26c31cc253.
>>> ..
>>> >
>>> > ?
>>> >
>>> > Or can the teams have different numbers of players?
>>> >
>>> > On May 30, 2:28 pm, Veer Sharma  wrote:
>>> >
>>> > > Hi Friends,
>>> >
>>> > > This is my first post to this forum. A "Hi" to all of you and here is
>>> > > my first problem...
>>> >
>>> > > Giiven int n, the total number of players and their skill-point.
>>> > > Distribute the players on 2 evenly balanced teams.
>>> >
>>> > > Lets see who gives the best solution (least space complexity / least
>>> > > time complexity or both...)
>>>
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Re: [algogeeks] Re: partion of array

[divya jain]

u have not sorted the array . first sort the array nd then apply the
approach. u ll get the ryt ans

On 1 June 2010 21:32, Jitendra Kushwaha  wrote:

> Using subset sum method we can solve this. It actually DP
>
> check out Paybill question and its solution on topcoder website
>
> link : http://www.topcoder.com/stat?c=problem_statement&pm=3114&rd=5865
>
> you will have a better understanding of subset sum problem after this
>
> --
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> Jitendra Kushwaha
> Undergradute Student
> Computer Science & Eng.
> MNNIT, Allahabad
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Re: [algogeeks] Implementing 2 stacks within a single linear array

[divya jain]

for three stacks u can use indices 0 3 6 etc for stack1. 1 4 7 for stack 2
and 2 5 8 etc for stack 3.
now if any of the stack overflows but there is still space in array as other
stacks have few element then now u can grow ur stack in reverse direction as
shown below :

indices of array :   0  1  2  3  4  5  6  7  8  9  10  11 12

contents  1   2  3  1 1  1  1
/  \
  \
  top2top3
   top1
here 1 represents element of stack1 2 for stack 2 and so on.
now if  u want to insert more element in stack 1 it will overflow while 2
and 3 have space so wat u can do now is grow stack 1 in reverse direction.
ie now place elements for stack 1 in index 11 and so now top1 will point to
11 and so on. u can indicate this behaviour of stack1 by using some tag.
i hope this will work..

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Re: [algogeeks] Check if 2 linked lists are identical

[divya jain]

construct bst from both list and check if the bst are equal by printing
their inorder traversal.

On 2 June 2010 22:47, Raj N  wrote:

> @Antony: The 2 lists should have the same elements as well the number must
> be equal
>
>
> On Wed, Jun 2, 2010 at 5:38 PM, Antony Vincent Pandian.S. <
> sant...@gmail.com> wrote:
>
>> @Raj
>>
>> What do you mean by identical? You are just concerned about the
>> presence of one element in another LL or you are concerned about the
>> equality of number of elements too?
>>
>> On 6/2/10, Raj N  wrote:
>> > @sharad kumar: But don't you think this'll consume a lot of space. Merge
>> > sort will give O(nlogn) complexity when a separate LL is used to store
>> the
>> > sorted elements but if we disbuild the same LL it'll be >O(n2). And also
>> wat
>> > do u mean by combining LL can u explain
>> >
>> >
>> > On Wed, Jun 2, 2010 at 6:48 AM, sharad kumar > >wrote:
>> >
>> >> @Raj:sorting can be done in O(nlogn)..sort both and compare both.or use
>> a
>> >> hash map to store all elements of one LL and then compare it with
>> >> otheror combine both the  LL perform merge sort and start deleting
>> >> adjacent elements.if adjacent elements in equal count then LL are equal
>> >> and
>> >> at end of process we get an empty list.
>> >>
>> >>
>> >> On Tue, Jun 1, 2010 at 11:55 PM, Raj N  wrote:
>> >>
>> >>> Hi all,
>> >>> Can someone suggest an efficient algorithm to check if 2 linked lists
>> >>> are identical. If 2 lists have to be sorted then there'll be a lot of
>> >>> space consumption for 2 separate linked lists. Can the whole process
>> >>> be done < O(n2)
>> >>>
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Re: [algogeeks] Re: string question

[divya jain]

the output shd be epo..

hint to the problem :   PROBLEM DO NOT READ IF U WANT TO SOLVE THE URSELF
it involves the concept of finding window
u hv to 1st search for the window which contains all characters of string.
then u have to alter window so as to get minmum length window..

On 17 May 2010 16:44, Modeling Expert  wrote:

> @Divya
> BigS =" Hellepo What's up"
> SmallS = 'eo'
> o/p should be ?  "ellepo" OR "epo" ?
>
> if its "ellepo" DP would work fine . If its "eo" probably need some
> modification in DP.
>
> -Manish
>
>
> On May 16, 8:36 pm, Navin Naidu  wrote:
> > @Sharad: yup
> >
> > On Sun, May 16, 2010 at 8:36 PM, Rohit Saraf <
> rohit.kumar.sa...@gmail.com>wrote:
> >
> >
> >
> > > @Navin: and that works ! :)
> > > @all : i am sure no heuristic/greedy strategy can be applied.
> > > @divya : did you check your array partitioning algorithm with my
> example !
> >
> > > --
> > > Rohit Saraf
> > > Second Year Undergraduate,
> > > Dept. of Computer Science and Engineering
> > > IIT Bombay
> > >http://www.cse.iitb.ac.in/~rohitfeb14
> 
> >
> > >  --
> > > You received this message because you are subscribed to the Google
> Groups
> > > "Algorithm Geeks" group.
> > > To post to this group, send email to algoge...@googlegroups.com.
> > > To unsubscribe from this group, send email to
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> 
> >
> > > .
> > > For more options, visit this group at
> > >http://groups.google.com/group/algogeeks?hl=en.
> >
> > --
> > Thanks & Regards,
> >
> > - NMN
> >
> > --
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Re: [algogeeks] Re: partion of array

[divya jain]

my algo on the array 1 200 500 2000
sort the array therefore we have now 2000 500 200 1
1st array will have largest element
A= {2000}
and B={500}
sumA=2000
sumB=500
now abs((2000+200)-500)>abs((2000)-(500+200))
so we ll put 200 in array B. now since B has n/2 elements rest of the
element goes to array which is 1.
so the ans is
A={2000,1}
b={500,200}


On 15 May 2010 19:10, Rohit Saraf  wrote:

> so what will ur algo give for array 1,200,500,2000
>
> On 5/15/10, divya jain  wrote:
> > my approach:
> > 1. sort the array
> > 2. take a variable diff. intitialize it to 0.
> > 3. take the 1st element from array nd place it in array A and 2nd element
> in
> > array B. stoe in diff sum(A)-sum(B).
> > 4. now place the next element in array A or B according to the condition
> if
> >if sum(A+element)-sum(B)> sum(a)-sum(B+element). store the element
> in
> > B  otherwise in A. also while storing the element in ny array maintain
> the
> > count of element in that aaray. if any time the count reaches n/2 where n
> is
> > the no. of elements in  the given aaray. then store rest element in the
> > other array.
> > 5. repeat step 5 until both array A n B get n/2 elements..
> >
> > hope my approach is clear and correct.
> > comments are welcomed.
> >
> > On 15 May 2010 08:47, Rohit Saraf  wrote:
> >
> >> Choosing a greedy strategy for this would be difficult.
> >>
> >> For a simple dp you can
> >> maintain A[i,total,present] using a recurrence
> >>
> >> i is the present index of array
> >> total is the number of elements reqd in first partition.
> >> present is the no of elements already there in first partition.
> >>
> >> the array stores difference between sums. GET the minimum of all these
> >> and backtrack.
> >>
> >>
> >> On 5/15/10, Amir hossein Shahriari 
> >> wrote:
> >> > @karas: your solution is greedy and its wrong e.g. for {1,2,3,4,5,100}
> >> your
> >> > diff would be 95 but the best result is 91
> >> >
> >> > i think we can solve this problem by dynamic programming but not a
> >> > simple
> >> > one! since the size of the two subsets must be equal.
> >> > so it's DP solution has at least 3 dimensions: tow dimensions
> >> representing
> >> > the number of elements in each subset and another for the difference
> >> between
> >> > their sums
> >> >
> >> > On Fri, May 14, 2010 at 10:11 PM, W Karas  wrote:
> >> >
> >> >> On May 14, 4:51 am, divya  wrote:
> >> >> > Algorithm to partition set of numbers into two s.t. diff bw their
> >> >> > sum is min and they hav equal num of elements
> >> >>
> >> >> void part(const int a[], int n_a, int g1[], int g2[])
> >> >>  {
> >> >>int i, j, k;
> >> >>
> >> >>/* diff = sum(g1) - sum(g2) */
> >> >>int diff;
> >> >>
> >> >>sort(a, n_a);
> >> >>
> >> >>diff = 0;
> >> >>for (i = 0, j = 1, k = 0; j < n_a; ++k, i += 2, j += 2)
> >> >>  {
> >> >>if ((a[i] > a[j]) == (diff > 0))
> >> >>  {
> >> >>g1[k] = a[j];
> >> >>g2[k] = a[i];
> >> >>  }
> >> >>else
> >> >>  {
> >> >>g1[k] = a[i];
> >> >>g2[k] = a[j];
> >> >>  }
> >> >>diff += g1[k] - g2[k];
> >> >>   }
> >> >>  }
> >> >>
> >> >> --
> >> >> You received this message because you are subscribed to the Google
> >> Groups
> >> >> "Algorithm Geeks" group.
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> >
> >> 
> 
> >
> >> >
> >> >> .
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> >> >> http://groups.google.com/group/algogeeks?hl=en.
> >> >>
> >> >>
> >> >
> >> > --
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> >> > Groups
> >> > "Algorithm

Re: [algogeeks] string question

[divya jain]

  output ll be : e's

On 16 May 2010 20:17, sharad kumar  wrote:

> suppose if i give
> Ssmall:es
> Sbig:he's  a  algogeek and he's rocking
> wat will be o/p?
>
>
> On Sun, May 16, 2010 at 8:12 PM, divya  wrote:
>
>> You r given a large string of characters lets call it Sbig. Then there
>> is a small set of characters lets call it Ssmall. You have to find
>> smallest substring of Sbig which contains all characters in Ssmall.
>>
>> For example you are given Sbig = "hello what are you doing"
>> Ssmall= "eo"
>> answer is "ello"
>>
>> --
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>> .
>> For more options, visit this group at
>> http://groups.google.com/group/algogeeks?hl=en.
>>
>>
>
>
> --
> yezhu malai vaasa venkataramana Govinda Govinda
>
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Re: [algogeeks] Re: partion of array

[divya jain]

my approach:
1. sort the array
2. take a variable diff. intitialize it to 0.
3. take the 1st element from array nd place it in array A and 2nd element in
array B. stoe in diff sum(A)-sum(B).
4. now place the next element in array A or B according to the condition if
   if sum(A+element)-sum(B)> sum(a)-sum(B+element). store the element in
B  otherwise in A. also while storing the element in ny array maintain the
count of element in that aaray. if any time the count reaches n/2 where n is
the no. of elements in  the given aaray. then store rest element in the
other array.
5. repeat step 5 until both array A n B get n/2 elements..

hope my approach is clear and correct.
comments are welcomed.

On 15 May 2010 08:47, Rohit Saraf  wrote:

> Choosing a greedy strategy for this would be difficult.
>
> For a simple dp you can
> maintain A[i,total,present] using a recurrence
>
> i is the present index of array
> total is the number of elements reqd in first partition.
> present is the no of elements already there in first partition.
>
> the array stores difference between sums. GET the minimum of all these
> and backtrack.
>
>
> On 5/15/10, Amir hossein Shahriari 
> wrote:
> > @karas: your solution is greedy and its wrong e.g. for {1,2,3,4,5,100}
> your
> > diff would be 95 but the best result is 91
> >
> > i think we can solve this problem by dynamic programming but not a simple
> > one! since the size of the two subsets must be equal.
> > so it's DP solution has at least 3 dimensions: tow dimensions
> representing
> > the number of elements in each subset and another for the difference
> between
> > their sums
> >
> > On Fri, May 14, 2010 at 10:11 PM, W Karas  wrote:
> >
> >> On May 14, 4:51 am, divya  wrote:
> >> > Algorithm to partition set of numbers into two s.t. diff bw their
> >> > sum is min and they hav equal num of elements
> >>
> >> void part(const int a[], int n_a, int g1[], int g2[])
> >>  {
> >>int i, j, k;
> >>
> >>/* diff = sum(g1) - sum(g2) */
> >>int diff;
> >>
> >>sort(a, n_a);
> >>
> >>diff = 0;
> >>for (i = 0, j = 1, k = 0; j < n_a; ++k, i += 2, j += 2)
> >>  {
> >>if ((a[i] > a[j]) == (diff > 0))
> >>  {
> >>g1[k] = a[j];
> >>g2[k] = a[i];
> >>  }
> >>else
> >>  {
> >>g1[k] = a[i];
> >>g2[k] = a[j];
> >>  }
> >>diff += g1[k] - g2[k];
> >>   }
> >>  }
> >>
> >> --
> >> You received this message because you are subscribed to the Google
> Groups
> >> "Algorithm Geeks" group.
> >> To post to this group, send email to algoge...@googlegroups.com.
> >> To unsubscribe from this group, send email to
> >> algogeeks+unsubscr...@googlegroups.com
> 
> >
> >> .
> >> For more options, visit this group at
> >> http://groups.google.com/group/algogeeks?hl=en.
> >>
> >>
> >
> > --
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> >
> >
>
>
> --
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
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Re: [algogeeks] Re: question

[divya jain]

slight change to the first algorithm.. if the element doesn't exist, then at
the end of the binary search, we'd have to look at the element on the left,
current, and right to see which one gives the closest result.

On 15 May 2010 16:35, divya jain  wrote:

> Very simple N^2*logN solution: sort the input array, then go through all
> pairs Ai, Aj (N^2 time), and for each pair check whether (S - Ai - Aj) is
> in array (logN time).
>
>
> On 12 May 2010 23:08, jalaj jaiswal  wrote:
>
>> @sateesh
>> suppose after sorting the array is
>> -99,-98,-97,-96,-95,-2,-1,0,4,5,99
>>
>> the answer should be {-99,0,99}.. sum is closest to zero here
>> so i dnt think the transition method works
>>
>>
>>
>>
>>
>>
>>
>>
>>
>>
>> On Fri, May 7, 2010 at 9:58 PM, sateesh  wrote:
>>
>>> I think this can be solved in better way.
>>>
>>> 1) Store the copy of array to get the indexes for the elements at the
>>> end of algo
>>> 2) Sort the array time: O(nlogn), space: O(1)
>>> 3) If first element of array is -ve and last element of array is not -
>>> ve, find the element at
>>>   which -ve to +ve transition happens
>>>   ex: -a -b +c +d
>>>   check the absolute minimum of following combinations and pick
>>> the correct pair
>>>-b+c+d
>>>-a+c+d
>>>-a-b+c
>>>-a-b+d
>>>here I assumed two +ve, two -ve. If only one -ve or one +v
>>> exists, we can pick the correct 3 elements straight away
>>>   else if all are -ve numbers , pick last 3 elements
>>>   else pick first 3 elements
>>>
>>>   This total process takes O(n) time
>>>  4) Based on picked three elements, do linear search on copied array
>>> to get there indexes.
>>>time: O(n)
>>>
>>> Overall it takes O(nlogn) time complexity and O(n) space complexity.
>>>
>>> Do you guys find any flaw in it ?
>>>
>>> -Sateesh
>>> IIT Kanpur 2004 M.Tech CSE Batch.
>>>
>>>
>>>
>>>
>>>
>>> On May 4, 10:43 pm, Afroz Mohiuddin  wrote:
>>> > Well if you want a sum of exactly 0 (or any constant) , there is an
>>> O(N^2)
>>> > way
>>> >
>>> > Take your array, and hash it, note that it is always possible to hash a
>>> > static set of keys so that the search/find in it is worst case O(1).
>>> This
>>> > takes O(N) space, and time.
>>> >
>>> > Then over all the tuples of numbers in the original array (a,b) check
>>> if 0 -
>>> > (a+b) is there in the hash set, time complexity O(N*N).
>>> >
>>> > For closest to 0 I guess the above solution is good.
>>> >
>>> > On Mon, May 3, 2010 at 2:18 PM, jalaj jaiswal <
>>> jalaj.jaiswa...@gmail.com>wrote:
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > > given an array(unsorted) may contain negative numbers too
>>> > > find the index of three numbers whose sum is closest to zero  in
>>>  O(N2 log
>>> > > N) time and O(N) space.
>>> >
>>> > > P.S -3 is more close to zero then -6 (number line ...)
>>> > > --
>>> > > With Regards,
>>> > > Jalaj Jaiswal
>>> > > +919026283397
>>> > > B.TECH IT
>>> > > IIIT ALLAHABAD
>>> >
>>> > > --
>>> > > You received this message because you are subscribed to the Google
>>> Groups
>>> > > "Algorithm Geeks" group.
>>> > > To post to this group, send email to algoge...@googlegroups.com.
>>> > > To unsubscribe from this group, send email to
>>> > > algogeeks+unsubscr...@googlegroups.com
>>> 
>>> >
>>> > > .
>>> > > For more options, visit this group at
>>> > >http://groups.google.com/group/algogeeks?hl=en.
>>> >
>>> > --
>>> > We are here on earth to do good for others. What the others are here
>>> for, I
>>> > don't know.
>>> >
>>> > Afroz Mohiuddin
>>> > Final Year Masters Student
>>> > Dept Computer Science and Engineering
>>> > Indian Institute of Technology Kanpur
>>> > Kanpur - 208016
>>> > INDIA
>>> >
>>> > --
>>> > You received this message b

Re: [algogeeks] Re: question

[divya jain]

Very simple N^2*logN solution: sort the input array, then go through all
pairs Ai, Aj (N^2 time), and for each pair check whether (S - Ai - Aj) is in
array (logN time).

On 12 May 2010 23:08, jalaj jaiswal  wrote:

> @sateesh
> suppose after sorting the array is
> -99,-98,-97,-96,-95,-2,-1,0,4,5,99
>
> the answer should be {-99,0,99}.. sum is closest to zero here
> so i dnt think the transition method works
>
>
>
>
>
>
>
>
>
>
> On Fri, May 7, 2010 at 9:58 PM, sateesh  wrote:
>
>> I think this can be solved in better way.
>>
>> 1) Store the copy of array to get the indexes for the elements at the
>> end of algo
>> 2) Sort the array time: O(nlogn), space: O(1)
>> 3) If first element of array is -ve and last element of array is not -
>> ve, find the element at
>>   which -ve to +ve transition happens
>>   ex: -a -b +c +d
>>   check the absolute minimum of following combinations and pick
>> the correct pair
>>-b+c+d
>>-a+c+d
>>-a-b+c
>>-a-b+d
>>here I assumed two +ve, two -ve. If only one -ve or one +v
>> exists, we can pick the correct 3 elements straight away
>>   else if all are -ve numbers , pick last 3 elements
>>   else pick first 3 elements
>>
>>   This total process takes O(n) time
>>  4) Based on picked three elements, do linear search on copied array
>> to get there indexes.
>>time: O(n)
>>
>> Overall it takes O(nlogn) time complexity and O(n) space complexity.
>>
>> Do you guys find any flaw in it ?
>>
>> -Sateesh
>> IIT Kanpur 2004 M.Tech CSE Batch.
>>
>>
>>
>>
>>
>> On May 4, 10:43 pm, Afroz Mohiuddin  wrote:
>> > Well if you want a sum of exactly 0 (or any constant) , there is an
>> O(N^2)
>> > way
>> >
>> > Take your array, and hash it, note that it is always possible to hash a
>> > static set of keys so that the search/find in it is worst case O(1).
>> This
>> > takes O(N) space, and time.
>> >
>> > Then over all the tuples of numbers in the original array (a,b) check if
>> 0 -
>> > (a+b) is there in the hash set, time complexity O(N*N).
>> >
>> > For closest to 0 I guess the above solution is good.
>> >
>> > On Mon, May 3, 2010 at 2:18 PM, jalaj jaiswal <
>> jalaj.jaiswa...@gmail.com>wrote:
>> >
>> >
>> >
>> >
>> >
>> > > given an array(unsorted) may contain negative numbers too
>> > > find the index of three numbers whose sum is closest to zero  in  O(N2
>> log
>> > > N) time and O(N) space.
>> >
>> > > P.S -3 is more close to zero then -6 (number line ...)
>> > > --
>> > > With Regards,
>> > > Jalaj Jaiswal
>> > > +919026283397
>> > > B.TECH IT
>> > > IIIT ALLAHABAD
>> >
>> > > --
>> > > You received this message because you are subscribed to the Google
>> Groups
>> > > "Algorithm Geeks" group.
>> > > To post to this group, send email to algoge...@googlegroups.com.
>> > > To unsubscribe from this group, send email to
>> > > algogeeks+unsubscr...@googlegroups.com
>> 
>> >
>> > > .
>> > > For more options, visit this group at
>> > >http://groups.google.com/group/algogeeks?hl=en.
>> >
>> > --
>> > We are here on earth to do good for others. What the others are here
>> for, I
>> > don't know.
>> >
>> > Afroz Mohiuddin
>> > Final Year Masters Student
>> > Dept Computer Science and Engineering
>> > Indian Institute of Technology Kanpur
>> > Kanpur - 208016
>> > INDIA
>> >
>> > --
>> > You received this message because you are subscribed to the Google
>> Groups "Algorithm Geeks" group.
>> > To post to this group, send email to algoge...@googlegroups.com.
>> > To unsubscribe from this group, send email to
>> algogeeks+unsubscr...@googlegroups.com
>> .
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>> groups.google.com/group/algogeeks?hl=en.
>>
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>>
>
>
> --
> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
>
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Re: [algogeeks] BST

[divya jain]

form a sorted  array from inorder traversal of tree. now take to pointers
one to the beginning of array and other at the end. now check if the sum of
element is greater than reqd sum then increment 1st ptr. if their sum is
less than reqd sum then decrement 2nd ptr. if their sum is equal to the reqd
sum then this is the ans..
hope it will work..

On 13 May 2010 20:11, jalaj jaiswal  wrote:

> given a bst... find two nodes whose sum is equal to a number k ... in O(n)
> time and constant space...
>
> --
> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
>
> --
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Re: [algogeeks] compute kth smallest element from union of two lists

[divya jain]

sorry bt i dint get the approach. can u please explain a bit more by taking
examples...thanks a lot in advance

On 14 May 2010 10:12, Rohit Saraf  wrote:

> This was also asked in my Yahoo! Interview this year. :)
>
>
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
>
> On Fri, May 14, 2010 at 10:10 AM, Rohit Saraf  > wrote:
>
>> exactly ..
>>
>> --
>> Rohit Saraf
>> Second Year Undergraduate,
>> Dept. of Computer Science and Engineering
>> IIT Bombay
>> http://www.cse.iitb.ac.in/~rohitfeb14
>>
>>
>>
>> On Fri, May 14, 2010 at 10:07 AM, vignesh radhakrishnan <
>> rvignesh1...@gmail.com> wrote:
>>
>>> This is for kth largest. Change it for kth smallest
>>>
>>> In fact, this problem is amenable to something very similar to binary
>>> search. Suppose my arrays are A and B. The idea is to keep track of two
>>> indices, a (for A) and b (for B), such that a + b = k - 1 (it's very
>>> important to maintain this invariant). It's easy to check whether A[a] or
>>> B[b] is the answer: A[a] is the answer if and only if
>>>
>>> B[b-1] <= A[a] <= B[b],
>>>
>>> and B[b] is the answer if and only if
>>>
>>> A[a-1] <= B[b] <= A[a],
>>>
>>> where we use the convention that A[-1] = B[-1] = "negative infinity."
>>> (This can be determined in constant time.) Moreover, if neither of these is
>>> the case, you can divide the problem in half: if A[a] < B[b-1], you restrict
>>> your attention to the portion of A above index a and the portion of B below
>>> index b, and otherwise (it must be the case that B[b] < A[a-1]), you
>>> restrict your attention to the portion of A below index a and the portion of
>>> B above index b. (If you start with a and b in the middle of the arrays and
>>> always make the new indices be in the middle of the subarrays you're
>>> considering at that point, this step will always divide the problem in
>>> half.)
>>> Thanks to Lux Perpetua 
>>> source:
>>>
>>> http://forums.devshed.com/software-design-43/finding-kth-largest-element-in-a-union-of-two-arrays-137322.html
>>>
>>>
>>> On 13 May 2010 19:34, divya  wrote:
>>>
 You are given two sorted lists of size m and n. Give an O(log m+log n)
 time algorithm for computing the kth smallest element in the union of
 the two lists

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Re: [algogeeks] frequency

[divya jain]

use binary tree and insert in it every character u come across. if the
character is already present then increment its count. use this approach if
u r nt sure that characters will be only 26 or no.
if u r sure there r 26 char then u cn use hash..
plz correct me if i m wrong.
thanks

On 14 May 2010 08:50, sharad kumar  wrote:

> cant u use a hash map  of O(K) where K is distinct elements in
> string..
>
>
> On Thu, May 13, 2010 at 8:13 PM, jalaj jaiswal 
> wrote:
>
>> input a character array from the user and output in the following way.
>> example string is amazon.. then output should be a2m1z1o1n1
>>
>> i did it taking a 26 size array... some better solution ??
>>
>>
>> --
>> With Regards,
>> Jalaj Jaiswal
>> +919026283397
>> B.TECH IT
>> IIIT ALLAHABAD
>>
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>
>
>
> --
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Re: [algogeeks] tree from linked list

[divya jain]

the best approach to it is to build a balanced tree from bottom to up rather
than top to bottom.

On 12 May 2010 22:47, divya jain  wrote:

> thanks a lot to all for their replies..
>
>
> On 9 May 2010 11:23, rahul rai  wrote:
>
>> can anyone give me links to more educative and active groups like
>> algogeeks
>>
>> On Sun, May 9, 2010 at 2:11 AM, Arun prasath
>>  wrote:
>> > This does not create a balanced tree but ensures that every element in
>> the
>> > tree is accessible by lg(n) time.
>> >
>> > Time : Complexity   O(n)
>> >
>> >
>> > [a...@91blore-srv1 ~]$ cat recursion.c
>> > #include 
>> > #include
>> > #include 
>> > #define TEST2
>> > #ifdef TEST1
>> > int arr[] = { 1,2,3,4,5,6,7};
>> > int max_elems = sizeof(arr)/sizeof(arr[0]);
>> > #endif
>> >
>> > #ifdef TEST2
>> > int arr[] = { 1,2,3,4,5};
>> > int max_elems = sizeof(arr)/sizeof(arr[0]);
>> > #endif
>> >
>> > #ifdef TEST3
>> > int arr[] = { 1,2,3,4,5,6,7,8};
>> > int max_elems = sizeof(arr)/sizeof(arr[0]);
>> > #endif
>> >
>> > #define LIST_EMPTY -1
>> >
>> > struct tree {
>> > int data;
>> > struct tree * left,* right;
>> > };
>> >
>> > struct tree* function( int , int);
>> > void print_inorder( struct tree *);
>> >
>> > int return_next_from_list(void)
>> > {
>> > static int nxt_elem = 0;
>> > if(nxt_elem < max_elems)
>> > return arr[nxt_elem++];
>> >
>> > return LIST_EMPTY;// empty condition
>> > }
>> > int main()
>> > {
>> > unsigned int  x = max_elems;
>> > struct tree* head;
>> > while( x & (x - 1) ) {
>> > x = x & (x - 1) ;
>> > }
>> >
>> > head = function(0, x);
>> > print_inorder(head);
>> > free(head);
>> > return 0;
>> > }
>> > struct tree* function(int mid, int i)
>> > {
>> > int val = mid + i ;
>> > if (val & 1) {
>> > struct tree * leaf = malloc( sizeof(struct tree) );
>> > leaf->left = leaf->right = NULL;
>> > leaf->data = return_next_from_list();
>> > if(leaf->data == LIST_EMPTY) {
>> > free(leaf);
>> > return NULL;
>> > }
>> > return leaf;
>> > }
>> > struct tree *non_leaf = malloc( sizeof(struct tree) ) ;
>> >
>> > non_leaf->left  = function( mid, i/2);
>> > non_leaf->data = return_next_from_list();
>> > if (non_leaf->data == LIST_EMPTY) {
>> > struct tree *tmp = non_leaf->left;
>> > free(non_leaf);
>> > return tmp;
>> > }
>> > non_leaf->right = function( mid+i, i/2);
>> > return non_leaf;
>> > }
>> > void print_inorder( struct tree* root)
>> > {
>> > struct tree * trav = root;
>> > if (!trav) {
>> > return;
>> > }
>> > print_inorder(trav->left);
>> > if(trav->left)
>> > free(trav->left);
>> > printf("{%d}", trav->data);
>> > print_inorder(trav->right);
>> > if(trav->right)
>> > free(trav->right);
>> >
>> > }
>> > [a...@91blore-srv1 ~]$
>> >
>> >
>> > On Sun, May 2, 2010 at 6:38 PM, divya  wrote:
>> >>
>> >> u are given a sorted lnked list construct a balanced binary search
>> >> tree from it.
>> >>
>> >> --
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>> >>
>> >
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Re: [algogeeks] 400!

[divya jain]

thanx to all 4 the solutions..

On 3 May 2010 18:39, Varun Nagpal  wrote:

> @Rajesh gave a simple elegant solution.
>
> A look at a Linux calculator : you can even calculate 99! =
> 8.854887824e+5584950 in few seconds. I just looked at the code(its open
> source right!), which is not so easy to understand in few minutes.
>
> Here is the some part of code I extracted from source files:
>
> /* Size of the multiple precision values */
> #define MP_SIZE 1000
>
> /* Base for numbers */
> #define MP_BASE 1
>
> /* Object for a high precision floating point number representation
>  *
>  * x = sign * (MP_BASE^(exponent-1) + MP_BASE^(exponent-2) + ...)
>  */
> typedef struct
> {
>/* Sign (+1, -1) or 0 for the value zero */
>int sign; //, im_sign;
>
>/* Exponent (to base MP_BASE) */
>int exponent; //, im_exponent;
>
>/* Normalized fraction */
>int fraction[MP_SIZE]; //, im_fraction[MP_SIZE];
> } MPNumber;
>
>
> void mp_factorial(const MPNumber *x, MPNumber *z)
> {
> int i, value;
>
> /* 0! == 1 */
> if (mp_is_zero(x)) {
> mp_set_from_integer(1, z);
> return;
> }
> if (!mp_is_natural(x)) {
> /* Translators: Error displayed when attempted take the factorial
> of a fractional number */
> mperr(_("Factorial is only defined for natural numbers"));
> mp_set_from_integer(0, z);
> return;
> }
>
> /* Convert to integer - if couldn't be converted then the factorial
> would be too big anyway */
> value = mp_cast_to_int(x);
> mp_set_from_mp(x, z);
> for (i = 2; i < value; i++)
> mp_multiply_integer(z, i, z);
> }
>
> mp_multiply_integer(z, i, z) subroutine is too big too put in here, too see
> its code visit:  http://live.gnome.org/Gcalctool
>  <http://live.gnome.org/Gcalctool>
> On Mon, May 3, 2010 at 2:34 PM, Anil C R  wrote:
>
>> @Jitendra
>> but that's no fun [?]
>>
>> -
>> Anil
>>
>>
>>
>> On Mon, May 3, 2010 at 5:12 PM, vignesh radhakrishnan <
>> rvignesh1...@gmail.com> wrote:
>>
>>> @siddharth and prasoon either design a very long integer library
>>> yourself, or use gmp library in cpp or BigInteger Class in java.
>>>
>>> Regards,
>>> vignesh
>>>
>>> On 3 May 2010 09:46, siddharth srivastava  wrote:
>>>
>>>> But is there any way to accomplish this without an array ? Even for
>>>> 100!.
>>>>
>>>>
>>>> On 2 May 2010 06:15, Prasoon Mishra  wrote:
>>>>
>>>>> I think challenge here is not the Execution time, but the storage. 300
>>>>> ! or 400! should generally go beyond the storage capabilities of long long
>>>>> ints in cpp.
>>>>> @ Rohit Saraf: Hence, I don't know if even tail recursion will
>>>>> ultimately be able to store the output.
>>>>> I think Rajesh Patidar's answer fits the bill well, in terms of
>>>>> storage.
>>>>>
>>>>>
>>>>> On Sun, May 2, 2010 at 2:23 PM, vignesh radhakrishnan <
>>>>> rvignesh1...@gmail.com> wrote:
>>>>>
>>>>>> I agree with abhijith. But given some very large x for which i would
>>>>>> have to find factorial.
>>>>>> I would either
>>>>>> (i) use gmp in cpp or BigInteger or java if its not a lab exercise or
>>>>>> an interview
>>>>>> (ii) use simple brute multiplication algorithm.
>>>>>> The second approach requires
>>>>>>  (a) The no. of digits in n! for making storage available
>>>>>>  (b) The calculation itself which I would brute force
>>>>>>
>>>>>> References:
>>>>>>
>>>>>> http://inder-gnu.blogspot.com/2009/08/find-number-of-digits-in-factorial-of.html
>>>>>>
>>>>>> http://stackoverflow.com/questions/1113167/can-one-know-how-large-a-factorial-would-be-before-calculating-it
>>>>>> http://delphiforfun.org/programs/big_factorials.htm
>>>>>>
>>>>>>
>>>>>>
>>>>>> On 2 May 2010 13:59, Rohit Saraf  wrote:
>>>>>>
>>>>>>> google it... u will gt it
>>>>>>>
>>>>>>> i am on mobile... cannot explain now..
>>>>>>>
>>>>>>> On 5/2/10, divya jain  wrote:
>>

Re: [algogeeks] tree from linked list

[divya jain]

thanks a lot to all for their replies..

On 9 May 2010 11:23, rahul rai  wrote:

> can anyone give me links to more educative and active groups like algogeeks
>
> On Sun, May 9, 2010 at 2:11 AM, Arun prasath
>  wrote:
> > This does not create a balanced tree but ensures that every element in
> the
> > tree is accessible by lg(n) time.
> >
> > Time : Complexity   O(n)
> >
> >
> > [a...@91blore-srv1 ~]$ cat recursion.c
> > #include 
> > #include
> > #include 
> > #define TEST2
> > #ifdef TEST1
> > int arr[] = { 1,2,3,4,5,6,7};
> > int max_elems = sizeof(arr)/sizeof(arr[0]);
> > #endif
> >
> > #ifdef TEST2
> > int arr[] = { 1,2,3,4,5};
> > int max_elems = sizeof(arr)/sizeof(arr[0]);
> > #endif
> >
> > #ifdef TEST3
> > int arr[] = { 1,2,3,4,5,6,7,8};
> > int max_elems = sizeof(arr)/sizeof(arr[0]);
> > #endif
> >
> > #define LIST_EMPTY -1
> >
> > struct tree {
> > int data;
> > struct tree * left,* right;
> > };
> >
> > struct tree* function( int , int);
> > void print_inorder( struct tree *);
> >
> > int return_next_from_list(void)
> > {
> > static int nxt_elem = 0;
> > if(nxt_elem < max_elems)
> > return arr[nxt_elem++];
> >
> > return LIST_EMPTY;// empty condition
> > }
> > int main()
> > {
> > unsigned int  x = max_elems;
> > struct tree* head;
> > while( x & (x - 1) ) {
> > x = x & (x - 1) ;
> > }
> >
> > head = function(0, x);
> > print_inorder(head);
> > free(head);
> > return 0;
> > }
> > struct tree* function(int mid, int i)
> > {
> > int val = mid + i ;
> > if (val & 1) {
> > struct tree * leaf = malloc( sizeof(struct tree) );
> > leaf->left = leaf->right = NULL;
> > leaf->data = return_next_from_list();
> > if(leaf->data == LIST_EMPTY) {
> > free(leaf);
> > return NULL;
> > }
> > return leaf;
> > }
> > struct tree *non_leaf = malloc( sizeof(struct tree) ) ;
> >
> > non_leaf->left  = function( mid, i/2);
> > non_leaf->data = return_next_from_list();
> > if (non_leaf->data == LIST_EMPTY) {
> > struct tree *tmp = non_leaf->left;
> > free(non_leaf);
> > return tmp;
> > }
> > non_leaf->right = function( mid+i, i/2);
> > return non_leaf;
> > }
> > void print_inorder( struct tree* root)
> > {
> > struct tree * trav = root;
> > if (!trav) {
> > return;
> > }
> > print_inorder(trav->left);
> > if(trav->left)
> > free(trav->left);
> > printf("{%d}", trav->data);
> > print_inorder(trav->right);
> > if(trav->right)
> > free(trav->right);
> >
> > }
> > [a...@91blore-srv1 ~]$
> >
> >
> > On Sun, May 2, 2010 at 6:38 PM, divya  wrote:
> >>
> >> u are given a sorted lnked list construct a balanced binary search
> >> tree from it.
> >>
> >> --
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> >>
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Re: [algogeeks] tree from linked list

[divya jain]

it is easy to traverse the array. as u can directly reach middle element in
constant time. but for linked list u cant do dat. so the time complexity of
both the solution given above are O(nlogn). is there any other better
complexity solution.

On 3 May 2010 17:41, jalaj jaiswal  wrote:

> for simplicity in writin algo i've taken sorted array instead of list
> struct node * create( int *sorted,number of elements){
>  struct node *temp,*left,*right;
>  int tempii[100],tempiii[100];
>  if(number of elemnts ==0)
>return NULL;
>  temp->data=sorted[mid];
>  temp->left=NULL;
>  temp->right=NULL;
>  if number of elements == 1
> return temp;
>  int count=0;
>  for(int i=0;i  tempii[i]=sorted[i];
>  count++;
>  }
>  left=create(int *tempii,count);
>  temp->left=left;
>  count=0;
>  for(i=mid+1;itempiii[i]=sorted[i];
>count++;
>  }
>  right=create(int *tempiii,count);
>  temp->right=right;
>
>  return temp;
>
> }
>
> On Mon, May 3, 2010 at 5:36 PM, Rohit Saraf 
> wrote:
>
>> 1) Make the middle element the root.
>>  Recursively make the left and right subtrees from the left and right
>> halves of the link list.
>>
>> 2) Implement balanced insertion in trees (via rotations on every step...).
>> Now insert each element
>> --
>> Rohit Saraf
>> Second Year Undergraduate,
>> Dept. of Computer Science and Engineering
>> IIT Bombay
>> http://www.cse.iitb.ac.in/~rohitfeb14
>>
>>
>>
>> On Sun, May 2, 2010 at 6:38 PM, divya  wrote:
>>
>>> u are given a sorted lnked list construct a balanced binary search
>>> tree from it.
>>>
>>> --
>>> You received this message because you are subscribed to the Google Groups
>>> "Algorithm Geeks" group.
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>>>
>>>
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>>
>
>
>
> --
> With Regards,
> Jalaj Jaiswal
> +919026283397
> B.TECH IT
> IIIT ALLAHABAD
>
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Re: [algogeeks] Re: a google question

[divya jain]

@satish

ur solution is of o(nlogn) complexty

@ jitendra

suppose p11 and p21 r pointing at index 0 and p12 at 4 and p22 at 1. now
suppose at ths point d s greater than b and c. now u increment p11 and p21.
but it can be a case that a[0] + b[2] is next greatest  value bt t wont
work for ur algo.

On 3 May 2010 00:46, Shishir Mittal <1987.shis...@gmail.com> wrote:

> @Algoose : No, it is not necessary that first n elements must contain A[0]
> or B[0]. For e.g.
> A = {40,30,15,10}
> B = {40,30,15,10}
>
> Required 4 largest values = {40+40, 40+30, 40+30, 30+30}.
>
>
> @Satish:
> A very nice algorithm provided by you.
> Complexity Analysis for the algorithm provided by you:
>
> Creation of max-heap of n elements = O(n).
> Time to add a new element to the heap after extracting the maximum - O(log
> n)
> Overall Complexity = O(n log n)
>
> Regards,
> Shishir
>
>
> On Sun, May 2, 2010 at 10:52 PM, Satish  wrote:
>
>> This problem can be simplified to a variation of merge part of merge
>> sort algorithm.
>>
>> - Break the set S having n^2 values into n individual arrays of length
>> n, say S1, S2,..Sn, where S1 = [a1+b1, a1+b2,...a1+bn].
>> - One can observe that each Si has entries which are themselves sorted
>> within Si.
>> - Perform merge of S1, S2,..Sn where take the largest values of each
>> Si, and create a heap of these individual max values.
>> - In each step, return the max value from heap and then add the next
>> max value from the Si that had the current max value and add it in the
>> max-heap.
>> - Repeat this step n times and then exit.
>>
>> Time: O(n).
>>
>> Satish
>>
>> On May 2, 5:41 pm, Algoose Chase  wrote:
>> > Hi
>> >
>> > will this work ?
>> >
>> > since we need Set S with n pairs of largest values , any such pair
>> within
>> > the set would (always) contain A[0] or B[0] because they maximize the
>> value
>> > of the pair.
>> >
>> > so
>> >
>> > // code snippet
>> > typedef std::vector Pairs;
>> >
>> > Pairs.push(A[0],B[0])
>> > int i = 1; // index for ListA
>> > int j = 1; // index for list B
>> > count = 1;
>> > while( count <= n)
>> > {
>> >   if( A[0] + B[j] > B[0] + A[i] )
>> >   {
>> > Pairs.push(A[0],B[j])
>> > j++;
>> >   }
>> >   else
>> >   {
>> >  Pairs.Add(A[i],B[0])
>> >  i++;
>> >   }
>> >   count++;
>> >
>> > }
>> >
>> > since there are n items in both the lists, j and i wont go beyond n in
>> the
>> > while loop
>> >
>> >
>
>
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Re: [algogeeks] 400!

[divya jain]

nice algo by rajesh.
bt i think using linked list will be better..

On 3 May 2010 09:46, siddharth srivastava  wrote:

> But is there any way to accomplish this without an array ? Even for 100!.
>
>
> On 2 May 2010 06:15, Prasoon Mishra  wrote:
>
>> I think challenge here is not the Execution time, but the storage. 300 !
>> or 400! should generally go beyond the storage capabilities of long long
>> ints in cpp.
>> @ Rohit Saraf: Hence, I don't know if even tail recursion will ultimately
>> be able to store the output.
>> I think Rajesh Patidar's answer fits the bill well, in terms of storage.
>>
>>
>> On Sun, May 2, 2010 at 2:23 PM, vignesh radhakrishnan <
>> rvignesh1...@gmail.com> wrote:
>>
>>> I agree with abhijith. But given some very large x for which i would have
>>> to find factorial.
>>> I would either
>>> (i) use gmp in cpp or BigInteger or java if its not a lab exercise or an
>>> interview
>>> (ii) use simple brute multiplication algorithm.
>>> The second approach requires
>>>  (a) The no. of digits in n! for making storage available
>>>  (b) The calculation itself which I would brute force
>>>
>>> References:
>>>
>>> http://inder-gnu.blogspot.com/2009/08/find-number-of-digits-in-factorial-of.html
>>>
>>> http://stackoverflow.com/questions/1113167/can-one-know-how-large-a-factorial-would-be-before-calculating-it
>>> http://delphiforfun.org/programs/big_factorials.htm
>>>
>>>
>>>
>>> On 2 May 2010 13:59, Rohit Saraf  wrote:
>>>
>>>> google it... u will gt it
>>>>
>>>> i am on mobile... cannot explain now..
>>>>
>>>> On 5/2/10, divya jain  wrote:
>>>> > wat is tail recursion plz explan in detail
>>>> >
>>>> > On 2 May 2010 08:15, Rohit Saraf  wrote:
>>>> >
>>>> >> @divya use tail recursion and rest should be fine..
>>>> >>
>>>> >> --
>>>> >> --
>>>> >> Rohit Saraf
>>>> >> Second Year Undergraduate,
>>>> >> Dept. of Computer Science and Engineering
>>>> >> IIT Bombay
>>>> >> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>>>> <http://www.cse.iitb.ac.in/%7Erohitfeb14>
>>>> >>
>>>> >> --
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>>>> Groups
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>>>> 
>>>> >
>>>> >> .
>>>> >> For more options, visit this group at
>>>> >> http://groups.google.com/group/algogeeks?hl=en.
>>>> >>
>>>> >>
>>>> >
>>>> > --
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>>>> Groups
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>>>> .
>>>> > For more options, visit this group at
>>>> > http://groups.google.com/group/algogeeks?hl=en.
>>>> >
>>>> >
>>>>
>>>>
>>>> --
>>>> --
>>>> Rohit Saraf
>>>> Second Year Undergraduate,
>>>> Dept. of Computer Science and Engineering
>>>> IIT Bombay
>>>> http://www.cse.iitb.ac.in/~rohitfeb14<http://www.cse.iitb.ac.in/%7Erohitfeb14>
>>>>
>>>> --
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>>>>
>>>>
>>>
>

Re: [algogeeks] a google question

[divya jain]

i found this question on some site and it was mentioned there todo in  o(n).
i dont have the solution
@ above

ur solution doesnt include the case of ncluding a[i] b[j] it takes only a[i]
b[0] or b[j] a[0]

On 2 May 2010 18:11, Algoose Chase  wrote:

> Hi
>
> will this work ?
>
> since we need Set S with n pairs of largest values , any such pair within
> the set would (always) contain A[0] or B[0] because they maximize the value
> of the pair.
>
> so
>
> // code snippet
> typedef std::vector Pairs;
>
> Pairs.push(A[0],B[0])
> int i = 1; // index for ListA
> int j = 1; // index for list B
> count = 1;
> while( count <= n)
> {
>   if( A[0] + B[j] > B[0] + A[i] )
>   {
> Pairs.push(A[0],B[j])
> j++;
>   }
>   else
>   {
>  Pairs.Add(A[i],B[0])
>  i++;
>   }
>   count++;
> }
>
> since there are n items in both the lists, j and i wont go beyond n in the
> while loop
>
>
> On Sun, May 2, 2010 at 3:44 PM, Shishir Mittal <1987.shis...@gmail.com>wrote:
>
>> This problem has been discussed before @
>> http://groups.google.co.in/group/algogeeks/browse_thread/thread/9c1e1aa8cf1ed437/d451cd9468d985f7
>>
>> I believe, the problem can't be solved in O(n) but only in O(nlog n).
>>
>> @Divya: Are you sure the interviewer explicitly asked for an O(n) time
>> algorithm?
>>
>>
>> On Sun, May 2, 2010 at 1:48 PM, vignesh radhakrishnan <
>> rvignesh1...@gmail.com> wrote:
>>
>>> @divya You're rite. Post a solution if you have one.
>>>
>>> --
>>> Regards,
>>> Vignesh
>>>
>>>
>>> On 2 May 2010 13:14, divya jain  wrote:
>>>
>>>> @Mohit
>>>>
>>>> according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
>>>> then i is incremented   i is now 2 so for next iteration u ll compare
>>>> a[2] b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever.
>>>> think for ths case also.
>>>>
>>>>
>>>> On 2 May 2010 11:22, mohit ranjan  wrote:
>>>>
>>>>> @Algoose Chase
>>>>>
>>>>> point taken
>>>>> Thanks
>>>>>
>>>>>
>>>>> Mohit Ranjan
>>>>> Samsung India Software Operations.
>>>>>
>>>>>
>>>>> On Sat, May 1, 2010 at 8:43 PM, Algoose Chase wrote:
>>>>>
>>>>>> @mohit
>>>>>>
>>>>>> The idea of DP is fine.
>>>>>> When you find the Max i dont think you need to include A[i+1]+B[j+1]
>>>>>> because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] 
>>>>>> since
>>>>>> both the lists are sorted in decreasing order.
>>>>>>
>>>>>>
>>>>>> On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan <
>>>>>> shoonya.mo...@gmail.com> wrote:
>>>>>>
>>>>>>> oops
>>>>>>>
>>>>>>> Sorry didn't read properly
>>>>>>> last algo was for array sorted in ascending order
>>>>>>>
>>>>>>> for this case, just reverse the process
>>>>>>>
>>>>>>>
>>>>>>> A[n] and B[n] are two array
>>>>>>>
>>>>>>> loop=n, i=0, j=0;
>>>>>>>
>>>>>>>
>>>>>>> while(loop>0)  // for n largest pairs
>>>>>>> {
>>>>>>>   print A[i]+B[j];  // sum of first index from both array
>>>>>>> will be max
>>>>>>>
>>>>>>>   foo = MAX ( A[i+1]+B[j], A[i+1]+B[j+1], A[i]+B[j+1] ) // using
>>>>>>> DP, moving forward
>>>>>>>
>>>>>>>   if foo==A[i+1]+B[j]; i++   // only increment A
>>>>>>>   if foo==A[i+1]+B[j+1]; i++; j++   // increment both A and B
>>>>>>>   if foo==A[i]+B[j+1]; j++  // increment only B
>>>>>>>
>>>>>>> }
>>>>>>>
>>>>>>>
>>>>>>>
>>>>>>> Mohit Ranjan
>>>>>>> Samsung India Software Operations.
>>>>>>>
>>>>>>>
>>>>>>> On Fri, Apr 30, 2010 at 8:40 PM, mohit ranjan <
>>>>>>> shoonya.mo...@gmail.com> wrote:
>>>>>

Re: [algogeeks] 400!

[divya jain]

wat is tail recursion plz explan in detail

On 2 May 2010 08:15, Rohit Saraf  wrote:

> @divya use tail recursion and rest should be fine..
>
> --
> --
> Rohit Saraf
> Second Year Undergraduate,
> Dept. of Computer Science and Engineering
> IIT Bombay
> http://www.cse.iitb.ac.in/~rohitfeb14
>
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>
>

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Re: [algogeeks] a google question

[divya jain]

@Mohit

according to ur algo if a[1], b[0] has sum greater than a[0],b[1]
then i is incremented   i is now 2 so for next iteration u ll compare a[2]
b[0] and a[1] b1]. but what about a[0] b[1] this pair s lost forever. think
for ths case also.

On 2 May 2010 11:22, mohit ranjan  wrote:

> @Algoose Chase
>
> point taken
> Thanks
>
>
> Mohit Ranjan
> Samsung India Software Operations.
>
>
> On Sat, May 1, 2010 at 8:43 PM, Algoose Chase wrote:
>
>> @mohit
>>
>> The idea of DP is fine.
>> When you find the Max i dont think you need to include A[i+1]+B[j+1]
>> because it can never be greater than both A[i+1]+B[j] and A[i]+B[j+1] since
>> both the lists are sorted in decreasing order.
>>
>>
>> On Fri, Apr 30, 2010 at 8:47 PM, mohit ranjan wrote:
>>
>>> oops
>>>
>>> Sorry didn't read properly
>>> last algo was for array sorted in ascending order
>>>
>>> for this case, just reverse the process
>>>
>>>
>>> A[n] and B[n] are two array
>>>
>>> loop=n, i=0, j=0;
>>>
>>>
>>> while(loop>0)  // for n largest pairs
>>> {
>>>   print A[i]+B[j];  // sum of first index from both array will be
>>> max
>>>
>>>   foo = MAX ( A[i+1]+B[j], A[i+1]+B[j+1], A[i]+B[j+1] ) // using DP,
>>> moving forward
>>>
>>>   if foo==A[i+1]+B[j]; i++   // only increment A
>>>   if foo==A[i+1]+B[j+1]; i++; j++   // increment both A and B
>>>   if foo==A[i]+B[j+1]; j++  // increment only B
>>>
>>> }
>>>
>>>
>>>
>>> Mohit Ranjan
>>> Samsung India Software Operations.
>>>
>>>
>>> On Fri, Apr 30, 2010 at 8:40 PM, mohit ranjan 
>>> wrote:
>>>
 Hi Divya,


 A[n] and B[n] are two array

 loop=n, i=n-1, j=n-1;

 while(loop>0)  // for n largest pairs
 {
   print A[i]+B[j];  // sum of last index from both array will be
 max

   foo = MAX ( A[i-1]+B[j], A[i-1]+B[j-1], A[i]+B[j-1] ) // using DP
 moving backward

   if foo=A[i-1]+B[j]; i--   // only reduce A
   if foo=A[i-1]+B[j-1]; i--; j--   // reduce both A and B
   if foo=A[i]+B[j-1]; j--  // reduce only B
 }

 Time: O(n)


 Mohit Ranjan



 On Fri, Apr 30, 2010 at 5:35 PM, divya wrote:

> Given two sorted postive integer arrays A(n) and B(n) (W.L.O.G, let's
> say they are decreasingly sorted), we define a set S = {(a,b) | a \in
> A
> and b \in B}. Obviously there are n^2 elements in S. The value of such
> a pair is defined as Val(a,b) = a + b. Now we want to get the n pairs
> from S with largest values. The tricky part is that we need an O(n)
> algorithm.
>
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