No distinction has been amongst stduents. I think it is abt incraesing the
distance between any two students.
On Sun, Jul 28, 2013 at 10:45 AM, Dave wrote:
> @Enchantress: I'm assuming that you are talking about cheating by copying
> from nearby students.
>
> If this is not the first exam, base
Hope this helps :
space: o(n^2)
time: o(n^2)
#include
using namespace std;
inline int max(int a,int b)
{
if(a>b)
return a;
else
return b;
}
int main()
{
char str[7]="hello";
int arr[3][3]={
{15,2,3},
{4,5,6},
# lengthy explanation give more attention
#here we are finding sums on all valid partition and storing all four
possible sums in variable a,b,c,d and and for all possible a,b,c,d we will
keep runninf max and min/
lets take an example parttion is done at row=0, coloumn=1
00 01| 02 03
@sahil Can you please explain your question with an example ?
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Navin , your reply is correct.
On Sat, May 19, 2012 at 10:36 PM, Gene wrote:
> The problem is not so clear, so you must make some assumptions to gat
> an answer. Since we have water, we have to envision the histogram in
> 3d. Then assume that the distance between histogram bars is 1 and bar
> i
i guess it can be done by modifying solution on
http://www.geeksforgeeks.org/archives/8405
my prev soln was based on the same..
instead of adding value to the stack...add index of that element.
in below code , line in bold are added
void nextSmaller(int arr[],int n)
{
s1 s;
int i,next,ele;
s.top=
Urm. It's probably not the same. We could find the maximum element in the
array and use the trivial approach till we reach the max_element. After
that, all we need to do is to shift all the elements right of max_element
to the left by 1 and place max_element at the end. But again..this isn't
O(n).
http://www.geeksforgeeks.org/archives/8405
^ Similar question.
On Sun, Mar 25, 2012 at 5:19 PM, atul anand wrote:
> wont work for all cases...ignore
> i will post the algoonce i fix it
> On 25 Mar 2012 17:06, "Amol Sharma" wrote:
>
>> @atul : it would be better for all to understand if you
wont work for all cases...ignore
i will post the algoonce i fix it
On 25 Mar 2012 17:06, "Amol Sharma" wrote:
> @atul : it would be better for all to understand if you write the algo
> instead of writing the code..
> --
>
>
> Amol Sharma
> Third Year Student
> Computer Science and Engineering
@shady : yes i guess this is what question says:-
so acc to this below algo work , i didnt execute it but i guess it will work
void nextSmaller(int arr[],int n)
{
s1 s;
int i,next,ele;
s.top=-1;
push(&s,0);
for(i=1;i next)
{
swap(arr,ele,i);
next=arr[ele];
if(isEmpty(&s)
@gene
i think for 3 4 2 you need to start from left most element, and then make
substitutions one by one.
so it will be
3 4 2
2 4 3
2 3 4
@all i googled a bit, and found that O(n) solution is possible for it, any
idea ?
On Sun, Mar 25, 2012 at 1:59 PM, Kartik Sachan wrote:
> +1 @saurabh...:P
>
+1 @saurabh...:P
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@amol I was trying to put forward the point that the o/p need not be
sorted.If you check the difference between time of my and payal's message
it was a case of race condition.
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Sun, Mar 25, 2012 at 6:54 AM, Ge
n = x%2 ?
x can be any integer.
On Fri, Dec 2, 2011 at 5:19 PM, Don wrote:
> (!x || !(x^1))
> !(x>>1)
> !((x|1)-1)
> (x*x)==x
> (x==(x==x))||(x==(x!=x))
>
> etc.
>
> On Nov 29, 9:07 pm, Nitin Garg wrote:
> > *What are the different ways to say, the value of x can be either a 0 or
> a
> > 1.*
>
Reverse elements of set from start to end
Reverse elements of set from end+1 to destination
Reverse elements of set from start to destination
DONE
O(n)
On Sat, Jul 9, 2011 at 7:25 PM, Yogesh Yadav wrote:
> @gopi: i didnt really understand what u want to say... what start,end and
> destination d
@gopi: i didnt really understand what u want to say... what start,end and
destination denotes here??
u said it should start with 1 but in result it is starting with 9...plz
explain ur question again
On Sat, Jul 9, 2011 at 7:21 PM, Gopi wrote:
> Hi Geeks
>
> Can anyone please comment on this
yes Heap Build is O(n)
but after build it will be nlgn for comparision. isn't it ?
On Tue, Jul 5, 2011 at 10:07 PM, vaibhav agarwal wrote:
> @Dave bt the heap build operation is O(n) there is a proof fr this
>
>
> On Tue, Jul 5, 2011 at 6:29 AM, saurabh singh wrote:
>
>> Yes I know I said it w
http://www.cim.mcgill.ca/~langer/250/2010/lecture24.pdf
On Tue, Jul 5, 2011 at 12:37 PM, vaibhav agarwal wrote:
> @Dave bt the heap build operation is O(n) there is a proof fr this
>
>
> On Tue, Jul 5, 2011 at 6:29 AM, saurabh singh wrote:
>
>> Yes I know I said it with regard to the current
@Dave bt the heap build operation is O(n) there is a proof fr this
On Tue, Jul 5, 2011 at 6:29 AM, saurabh singh wrote:
> Yes I know I said it with regard to the current problem
>
> On Tue, Jul 5, 2011 at 8:58 AM, Dave wrote:
>
>> @Saurabh: Nope. You can construct a heap in-place. But it is no
Yes I know I said it with regard to the current problem
On Tue, Jul 5, 2011 at 8:58 AM, Dave wrote:
> @Saurabh: Nope. You can construct a heap in-place. But it is not O(n).
>
> Dave
>
> On Jul 4, 10:02 pm, saurabh singh wrote:
> > Again heap will require extra space.
> >
> > On Tue, Jul 5, 2011
@saurabh bt we need only one extra array
On Mon, Jul 4, 2011 at 11:02 PM, saurabh singh wrote:
> Again heap will require extra space.
>
>
> On Tue, Jul 5, 2011 at 8:25 AM, vaibhav agarwal <
> vibhu.bitspil...@gmail.com> wrote:
>
>> what abt this...
>> check length of the array if same then we m
Again heap will require extra space.
On Tue, Jul 5, 2011 at 8:25 AM, vaibhav agarwal
wrote:
> what abt this...
> check length of the array if same then we make a min heap of both the
> arrays which can be done in O(n) and call extraxtmin(). in this way we can
> find whether they r equal.
> othwe
what abt this...
check length of the array if same then we make a min heap of both the
arrays which can be done in O(n) and call extraxtmin(). in this way we can
find whether they r equal.
othwersie nt equal.
correct me if i am wrong!!
On Mon, Jul 4, 2011 at 4:35 AM, saurabh singh wrote:
> Let
Lets conclude this post.Shall we?
.An o(n) seems infeasible without any significant extra memory
If extra memory is allowed,hash maps can be used to bring it down to
o(logn).But hash maps would eat up serious memory if numbers occupy a large
range.
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I guess this question is similar to anagram.
On Mon, Jul 4, 2011 at 1:07 AM, Arpit Sood wrote:
> Hey,
> what is the solution with XOR, methods mentioned above seem to
> fail there any reference ?
>
>
> On Sun, Jul 3, 2011 at 10:39 PM, Deoki Nandan wrote:
>
>> there is no possibl
Hey,
what is the solution with XOR, methods mentioned above seem to
fail there any reference ?
On Sun, Jul 3, 2011 at 10:39 PM, Deoki Nandan wrote:
> there is no possible solution for this question in less than O(nlgn) time.
> As by theorem given in cormen and solution is possib
may be we can add condition that sum of squares and cubes be same.
On Sun, Jul 3, 2011 at 1:09 PM, Deoki Nandan wrote:
> there is no possible solution for this question in less than O(nlgn) time.
> As by theorem given in cormen and solution is possible using xor
>
>
> On Sun, Jul 3, 2011 at 2:
there is no possible solution for this question in less than O(nlgn) time.
As by theorem given in cormen and solution is possible using xor
On Sun, Jul 3, 2011 at 2:27 PM, Sandeep Jain wrote:
> For case1) yes XOR works,
> for Well, for the other two cases hash-maps may come in handy. :)
>
>
> R
For case1) yes XOR works,
for Well, for the other two cases hash-maps may come in handy. :)
Regards,
Sandeep Jain
On Sun, Jul 3, 2011 at 1:48 PM, sunny agrawal wrote:
> But i don't think xor method will work at all for all of the cases above
> you mentioned.
> setA = {4,7}
> setB = {5,6}
>
>
But i don't think xor method will work at all for all of the cases above you
mentioned.
setA = {4,7}
setB = {5,6}
-> all numbers in both set are nonzero and distinct
-> all numbers are in some range :D
and for character parts it will similarly failby taking character set of
ascii values 4,5,6,
Agreed, BUT if you don't add a stipulation. You won't be able to reduce the
complexity.
For a 100% general solution, I don't think you can reduce the complexity
more than O(nLgn.)
There are variations of this question:
--> All numbers are non-zero and distinct.
--> All numbers belong to given range
@sandeep
SET A -> {0,3,4,7}
SET B -> {1,2,5,6}
xor of all elements is zero
sum of both the sets is same
no of elements in both are same
overall result : all Algorithm posted above Fails
On Sun, Jul 3, 2011 at 12:59 PM, Sandeep Jain wrote:
> I was thinking the same, BUT here the question is tha
I was thinking the same, BUT here the question is that we have two *SETS*
and that's the catch.
So, XORing all elements of SET A with SET B should result in ZERO only when
both the set have same elements.
Regards,
Sandeep Jain
On Sun, Jul 3, 2011 at 11:25 AM, Pranav Agarwal wrote:
> I think
I think that the above algo will fail for the following two arrays:
a={2,2,3,3}
b={4,4,1,1}
sum(a)=sum(b);
a^b=0;
len(a)=len(b);
Correct me if i am wrong!
Pranav
On Sun, Jul 3, 2011 at 7:43 AM, varun pahwa wrote:
> @aditya. xor all elements mean that. take xor of each element of 1st array
> st
@aditya. xor all elements mean that. take xor of each element of 1st array
store in a variable that take xor of variable and each element of the second
array if all elements are common then the variable will be 0 some where.
var = a[0];
for(i = 1; i < sizeof(a)/sizeof(a[0]); i++)
var = var ^ a[i];
@mohit..:i dint get the logic behind XOR plz explain ..nd ya i dont think
dat you can find second largest in less than O(n).
On Sun, Jul 3, 2011 at 2:43 AM, mohit mittal wrote:
> Dont think that the corresponding elements should be same.
> XOR Should do it anyway.
>
> Btw other question "How wou
Dont think that the corresponding elements should be same.
XOR Should do it anyway.
Btw other question "How would you find the second largest element in an
array using minimum no of comparisons?Any thing better than O(n)."?
On Sun, Jul 3, 2011 at 2:41 AM, aditya kumar
wrote:
> xor will only res
xor will only result if corresponding elements are same . what if in both
the array set of integers are same but they arnt corresponding to each other
??
On Sun, Jul 3, 2011 at 2:37 AM, Dumanshu wrote:
> xor all the elements of both arrays ==0
> sum of 1st array == sum of 2nd array
> no. of elem
but according to the question,ptr is pointing to the second node in this
case
On Fri, Apr 8, 2011 at 8:55 PM, Anurag atri wrote:
> if innitially temp is pointing to A then there is no problem in deleting
> the middle node ..
>
>
> On Fri, Apr 8, 2011 at 4:49 PM, murthy.krishn...@gmail.com <
>
if innitially temp is pointing to A then there is no problem in deleting the
middle node ..
On Fri, Apr 8, 2011 at 4:49 PM, murthy.krishn...@gmail.com <
murthy.krishn...@gmail.com> wrote:
> hii,
>
> Small correction
>
>
> For the second case,
>
> Consider,
>
> A -> B -> C -> NULL
>
> Initially te
hii,
Small correction
For the second case,
Consider,
A -> B -> C -> NULL
Initially temp is pointing to A.
Accor 2 me he has asked to reverse d list to make it as C -> A by deleting
B, which can be done like this,
temp->next = temp->next->next; // A->C->NULL
temp->next->next = temp; //A->C->A
For the second case,
Consider,
A -> B -> C -> NULL
Accor 2 me he has asked to reverse d list to make it as C -> A by deleting
B, which can be done like this,
temp->next = temp->next->next; // A->C->NULL
temp->next->next = temp; //A->C->A
temp = temp->next; //C->A->C
temp->next->next = NULL; //C
Why??? It doesn't help to solve problem. You are already have tree structure
with parent links. Taunt.
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On Tue, Jan 4, 2011 at 8:13 AM, rahul patil
wrote:
>
>
> On Mon, Jan 3, 2011 at 6:08 PM, juver++ wrote:
>
>> Tree structure already have parent node link. Even we reconstruct the tree
>> as linked list we are not allowed to achieve
>
>
> Normal tree node does not contain link to its parent. I am
On Mon, Jan 3, 2011 at 6:08 PM, juver++ wrote:
> Tree structure already have parent node link. Even we reconstruct the tree
> as linked list we are not allowed to achieve
Normal tree node does not contain link to its parent. I am not saying
convert tree into linklist directly. I want to say tha
Tree structure already have parent node link. Even we reconstruct the tree
as linked list we are not allowed to achieve the goal. Path can be combined
using non-contigious (created from inorder traversal) elements. The only
solution is using DP with O(MAX_SUM_VALUE) extra space for each node.
-
On Sun, Jan 2, 2011 at 8:30 PM, Akash Agrawal wrote:
> I have written a kinda messed-up code for the same. Which is basically a
> bottom-up approach.
>
> Please find the same as attached. Some boundary conditions might be missed
> and code can be written in a more decorated, beautiful fashion.
>
>
I have written a kinda messed-up code for the same. Which is basically a
bottom-up approach.
Please find the same as attached. Some boundary conditions might be missed
and code can be written in a more decorated, beautiful fashion.
Logic:
- start from the root,
- keep the nodes value in an
No , we had to find all the paths . Some paths could include the root .
On Tue, Dec 28, 2010 at 11:12 PM, yq Zhang wrote:
> I think the original question says "Path can go from left subtree tree ,
> include root and go to right tree as well". This should mean the path must
> include the root.
>
I think the original question says "Path can go from left subtree tree ,
include root and go to right tree as well". This should mean the path must
include the root.
On Tue, Dec 28, 2010 at 4:52 AM, shanushaan wrote:
> Not clear what path you are referring to.
>
> Question. Should the path includ
Check it once again. There is no submatrix with 8 1's in it.
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Hi ,
the program outputs the co-ordinate of bottom-right of the largest
1*1 rectangular submatrix and the size is total number of elements in that
matrix.
On Mon, Dec 27, 2010 at 7:31 PM, juver++ wrote:
> Program is incorrect. Why does it output the following answer: point at
> (3,5 )s
Program is incorrect. Why does it output the following answer: point at (3,5
)size is 8???
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Hi Guys ,
I have coded the first part soon i will come with the
solution of part 2 :---
Let me know if u find any error case (hope u will find none)
public class Largestsubmatrix {
public point [][] a;
int [][] binmatrix;
public point loc;
public Largestsubmatrix(int [][] a) {
t
How to solve the second question? it is different from the other question
posted where it requres only SQUARE sub matrix.
Sent from Nexus one
On Dec 25, 2010 11:00 AM, "juver++" wrote:
> Try to search the answer before sumbitting the question here.
>
> --
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@ritesh..you dnt have to output v.. you have to output the minimum number of
flips so that your tree evaluates to v(v is either 0 or 1)
and if it alreday evaluates to v then return 0(no flips required)
if not possible return -1
On Wed, Aug 4, 2010 at 12:11 AM, RITESH SRIVASTAV
wrote:
> level of
I hope the value of V is 0 or 1. Is this right?
On Wed, Aug 4, 2010 at 12:48 AM, Manjunath Manohar wrote:
> @above: i have little difficulty in perceiving the question... can u give
> certain test cases..sample input/output ..
>
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@above: i have little difficulty in perceiving the question... can u give
certain test cases..sample input/output ..
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write a recursive function getmin(node, value) that returns the least number
of flips necessary for the subtree rooted at "node" to give the result
"value". recursive relations are easy to come up with, so I leave it as an
exercise :)
memorize the values calculated, so, never calculate a result mo
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