3:51 AM, Don wrote:
> I believe that this will generate a maze with multiple cycles, which
> violates the requirement stated in the initial question that the maze
> have exactly one solution.
>
> On Feb 6, 11:53 am, Anup Ghatage wrote:
> > There is another algorithm.. T
ve the logic of your rnd() function?
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Correction: Upon the frequency of an element it's BROUGHT UP closer to the
root.
On Wed, Nov 30, 2011 at 6:39 AM, Anup Ghatage wrote:
> Splay Trees are self modifying trees.
> Upon the frequency of an element it's not is brought up closer to the root.
>
> So in a cont
umar raja wrote:
> @Anup:
>
> How splay trees come into picture here?? I think TRIE or Hashtable will
> work out.
>
> But it is difficult to conceive the TRIE solutions and explain it to the
> Interviewer .so i am looking for something else better and concrete one.
>
>
> O
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>> > Kumar Raja
>> > M.Tech(SIT)
>> > IIT Kharagpur,
>> > 10it60...@iitkgp.ac.in
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stored in sorted order. Get
> a new element from array and lets call this number as m. Push m into the
> stack. Now, find all elements which are <= (m-1) using binary search. Pop
> out all these elements and assign the value m in the output array. Elements
> remaining at the end will
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>> > > > > > > and use the well known O(N) median selection algorithm to get
>> a median
>> > > > > > > of this. This has the quality we want of being somewhere
>> roughly in
>> > > > > > > the middle half of the remaini
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> On Sat, Sep 17, 2011 at 8:30 AM, Anup Ghatage wrote:
>
>> As WgpShashank once pointed out.
>>
>> Search the whole matrix for the first character instances, for each
>> instance, send the array, string and that char's position to a function that
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5}
>> a
>>&a
>> a+1
>> &a+1
>> &a is address of whole array...so &a+1 means next element after array
>> completion.
>>
>> so ptr-1 will point to 5
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#include
int main()
{
int a[5]={1,2,3,4,5};
int *ptr=(&a + 1);
printf("%d %d\n",*(a+1),*(ptr-1));
return 0;
}
Find the output!
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Here is another one. Pardon me if it goes by some other name.
Divide the array in K parts.
Find the Maximum in each of these parts and store it in an array of size K.
Find the Minimum in K to find the K'th max.
Total Time Complexity = O ( n + k ) where 0 <= k <= n
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You can quite easily map this problem to that of a typical client-server
protocol.
You don't know the size of the data, and the all packets will be of a
certain size, possibly, except the last one.
So the solutions provided for that problem might apply to some cases of
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> NIT ALLAHABAD
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le , we don't give the size of
>> the table ahead ... It decides some size, and implements.. when we cross the
>> assumed size,then it again gives some more memory so that all the operations
>> are optimum (less collisions).
>>
>>
>> On Sun, Sep 4, 2011 a
2 in worst case.
So the algorithm would be O( m^2 log m ) just to populate the hash and find
the frequencies.
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That's interesting.
when n = 3
We have been given this :F F F | E E E
Swap the middle element and it becomes: F E F | E F E
Which is what you want, but when n = 5
F F F F F | E E E E E
And you swap the middle element it becomes: F F E F F | E E F E E
So the same startegy doesn't apply.
But
Just a quick note to the above solution.
If the tree is a binary search tree and you are going to put it in an array
after an inorder traversal, the elements will be sorted, so to find the k'th
max just access the n-k'th element
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Could you please give an example for question 3?
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What is the purpose of this elitmus test?
Does scoring good in it qualify you for something?
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I had once read somewhere that they had used gets() during the first
compilations of unix, and it used to always crash for some test cases.
So after that they stopped using it and alerted everyone of the problem..
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Sort the given array first. : O(n logn)
You'll have:
{2,3,3,5,5,5}
First parse all the elements in the array and identify the unique ones and
store the count in N
Now create a two dimensional array A[N][2];
Store the number in A[0][0] and store its frequency in A[0][1] etc.
: O (n)
Now, reverse
^ +1
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Isn't it obvious now?
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In other words, The original array identifer, and hence the pointer, is a
constant pointer.
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See, a is defined in main() therefore its scope is limited to main(). It is
stored probably in main()'s stack
When you pass it to a function as a pointer. You don't make a copy of 'a'
onto the function's stack.
The only way you can view, manipulate it is using a pointer, and that
pointer is your on
I just checked Shashank's blog post. The Deque solution is awesome :)
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Given an unsorted Array A and any integer k where k <= size of A
Print the maximum of each sub-array of size k of A.
eg: A = [ 3, 5, 1, 9, 0, 4, -1, 7 ] k = 4
Max: 9 9 9 9 7
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How about we go column-wise first.
Search for a '1' in each row, if found, make it a '2' and all the elements
below that, in that column '1'
1 0 0 1 2 0 0 2
0 0 1 0 > 1 0 2 1
0 0 0 0 1 0 1 1
Now search for a '2' in each row and if it is present make each element in
that row except
So, in short, find all permutations for a given string and create a linked
list from them?
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Anyone interested in laying out the trade off's of the Babylonian method vs
Newton Raphson?
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Possibly a case of finding the longest common sub-sequence and then doing
the insertions/deletions etc
or AKA Levenshtein distance / Edit Distance
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@Dave Your algo's average case working should be better than a naive O(n)..
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@tech
Could you give an example for this?
If I've understood the question correctly, if we are not allowed duplicates
in the array the array turns out to be sorted.
If we are allowed duplicates, do we return the first occurrence ?
@Dave
What would be the complexity of your solution?
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photo/my-images/202/treeexpl.png/
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@Rajeev:
I had written down the recursion tree on paper but since you said that there
were 10 reds and greens, I compiled the code and checked it myself.
I'm afraid you are wrong my friend.
The correct answer is indeed 6 RED and 10 GREEN calls.
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If I'm not mistaken,
red() will be called 6 times and green() will be called 10 times.
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Check this out:
Tie it at the 200th meter mark. Throw the 150mt rope down.
Climb down to the 100th meter pole. Tie the rope there from the middle, and
not the end.
So what you have is a 150 mt rope that is tied at 200 mt mark, 100 mt mark
and 50 mts of the rope from 100 mt marks is hanging.
Now
On Thu, Aug 25, 2011 at 8:06 PM, Shravan Kumar wrote:
> @Anup
>
> Can you please elaborate why it will not work?
>
>
> On Thu, Aug 25, 2011 at 8:02 PM, Anup Ghatage wrote:
>
>> Naren:
>> It wont work for tree's of level > 3 and non-complete binary tre
25.
>
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Hey, Tell me if this assumption of the problem statement is correct..
String: ram
Matrix:
|r | a b
|a | g h
|m| j d
This is vertical and similar for its horizontal counter part...
So the logic is more than obvious,
For vertical (and horizontal), do a sub-string search column-wise and
(row-
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Actually, this method will be O(n) for any number of occurrences of a single
word, but It will go into O(n^2) for multiple occurrences of multiple words.
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How about this,
We compare the first character with every character except the characters in
the word itself, if there is a hit, it might be a word which is a duplicate.
So from there on, compare each of the following characters till there is a
white space or new line character.
If there is no matc
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