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Thanking You
Ashish
, Complexity is O(N^4)
On 28 April 2010 13:36, Ashish Mishra amishra@gmail.com wrote:
you are given a M x N matrix with 0's and 1's
find the matrix with largest number of 1,
1. find the largest square matrix with 1's
2. Find the largest rectangular matrix with 1's
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One of my friend ask me this n i am bad in P C (will love if smone can
provide me a link to learn it though)
nyways prob is:
there are infy color balls of k different color
you are allowed to pick n balls out of those infy(infinite) balls
cond is : you must have all k color balls with u
you are given a M x N matrix with 0's and 1's
find the matrix with largest number of 1,
1. find the largest square matrix with 1's
2. Find the largest rectangular matrix with 1's
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into bst no need to use extra space u have
to just ditach the node from binary tree and attach it in bst.
On Wed, Apr 28, 2010 at 1:18 AM, Ashish Mishra amishra@gmail.com
wrote:
How to build BST from binary tree in place i.e without extra space ??
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Ashish
y u need backtracking
i think it can be done with DP
On Sat, Apr 10, 2010 at 9:12 AM, «« ÄÑÜJ »» anujlove...@gmail.com wrote:
Need help in designing efficient backtracking algorithm for the coin
changing problem. where a1, a2, an are the set of distinct coins
types, where each ai is an
i think u mean lowest commen ancestor?
On Wed, Apr 7, 2010 at 10:34 PM, Himanshu Aggarwal
lkml.himan...@gmail.comwrote:
For a given binary tree, given the root and two node pointers, how can we
find their youngest common ancestor.
Say the node is like:
struct node{
int data;
yup atul algo is correct
(cur_data - node-right) * ( cur_data- node-left) -1 for ancestors
On Thu, Apr 8, 2010 at 10:19 AM, atul verma atul.ii...@gmail.com wrote:
Its very simple to solve this.
Start from root.
Compare the value of current node data value to both nodes.
1. if both are
@ankur how u can solve it in o(n)
i suppose u need atleast o(n lgn)
On Sun, Sep 6, 2009 at 2:52 PM, ankur aggarwal ankur.mast@gmail.comwrote:
o(n) is the best sol known to me..
On Sun, Sep 6, 2009 at 1:54 PM, Pramod Negi negi.1...@gmail.com wrote:
i guess sorting will do the work.
any
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