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 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 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 the right of column X.
>
> > Return the minimum number of row swaps you need to achieve the goal.
>
> > Input
>
> > The first line of input gives the number of cases, T. T test cases follow.
> > The first line of each test case has one integer, N. Each of the next N
> > lines contains N characters. Each character is either 0 or 1.
>
> > Output
>
> > For each test case, output
>
> > Case #X: K
> > where X is the test case number, starting from 1, and K is the minimum
> > number of row swaps needed to have all the 1 values in the matrix below or
> > on the main
>
> > diagonal.
>
> > You are guaranteed that there is a solution for each test case.
>
> > Limits
>
> > 1 ≤ T ≤ 60
>
> > 1 ≤ N ≤ 8
>
> > Input
>
> > 3
> > 2
> > 10
> > 11
> > 3
> > 001
> > 100
> > 010
> > 4
> > 1110
> > 1100
> > 1100
> > 1000
> > Output
> > Case #1: 0
> > Case #2: 2
> > Case #3: 4
>
> > --
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