Its O(n*n) algo.. so you have to just iterate n times.. please take an
exampla and solve it.. u'll get to know urself :)
On Aug 19, 3:14 am, Prakash D cegprak...@gmail.com wrote:
hey, thanks..
but if it needs many iteration, then we've to check each time whether the
array is sorted.. is there
hey, thanks..
but if it needs many iteration, then we've to check each time whether the
array is sorted.. is there any better way for swapping
On Thu, Aug 18, 2011 at 5:02 AM, Brijesh Upadhyay
brijeshupadhyay...@gmail.com wrote:
IT is the question..
You are given an N x N matrix with 0 and 1
Make an array of size n, and fill it with index of last 1 of corresponding
row in the matrix.. as if row 0 has 1101100 , than arr[0]=4 .
so we end up with an array containing last position of 1 in the respective
row..
lets arr[4] is like 3,2,2,1... we have to make it 1,2,2,3.. so now apply
IT is the question..
You are given an N x N matrix with 0 and 1 values. You can swap any two
adjacent rows of the matrix.
Your goal is to have all the 1 values in the matrix below or on the main
diagonal. That is, for each X where 1 ≤ X ≤ N, there must be no 1 values in
row X
that are to
