@Rajesh Patidar
I think we should do in Post order traversal alone. If we go by
Preorder/Inorder we might lose track of children node that is currently
being inserted into the BST. - correct me if im wrong :)
On 28 April 2010 15:30, Rajesh Patidar patidarc...@gmail.com wrote:
pickup node in
Let Memo[i][j] be the sum of elements in the (sub) rectangle from (0,0) to
(i,j)
Then use principle of inclusion and exclusion to find the sum of elements
from (a, b) to (c, d) in O(1)
for N*N matrix, Complexity is O(N^4)
On 28 April 2010 13:36, Ashish Mishra amishra@gmail.com wrote:
you
@chitta koushik
No, That might lead to negative edge cycles!
2010/1/15 chitta koushik koushik.infin...@gmail.com
How abt negating values and using same single source shortest path algo ?
2010/1/15 saltycookie saltycoo...@gmail.com
longest path is NP-hard
2010/1/11 Johan
to maximize the minimum distance between any two points:- to maximize the
minimum distance between adjacent points
- for this all points must be equally spaced.
hence, choose 'n' equally spaced points in the range (r1, r2) starting from
r1 and ending at r2.
2009/10/21 saltycookie
hope you mean find *sum* of all the digits in number-100! :)
2009/10/11 vicky mehta...@gmail.com
find some of all the digits in number-100!
--
Reduce, Reuse and Recycle
Regards,
Vivek.S
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@Varun S VIt wont work for this 1 3 4 5
2009/9/2 Varun S V varun...@gmail.com
Since we difference between two minumum elements should suffice, how about
finding the min and second minimum element in the array in single scan and
returning their difference. This should take not more than O(N)
let A[i+1] = a[0] + a[1] + ... + a[i];
let B[i+1] = b[0] + b[1] + ... + b[i];
A[0] = B[0] = 0;
sum of nos from i to j of a[] = A[j+1] - A[i]; // O(1) time
Thus O(n^2) sol can be obtained for this problem by finding the sub-array
sum in the range [i, j] for every i, j;
But I have not used the fact
@Devi;
Consider this,
N = 2
The array elements should these (N+2), numbers : {1, 2, 3, 4}
If the given array is {4, 3}
Will your code work correctly?
2009/7/31 Devi G devs...@gmail.com
The logic is actually simple. Tot if we mark in some way an element when
it's scanned, we can find the
N! overflows...
Try to write a program to find the value of 30!
You don't have a variable that is large enough to store such a big number...
2009/7/31 sharad kumar aryansmit3...@gmail.com
check this out
Let x and y be the missing number,
Now equation 1 is : x + y = [n(n+1)/2] - S
equation
