@Anshu
If you do
" add top bottom, left right element to the popped element in qeuue "
should you need to do it for each element in the matrix.
So, will that not be O(n3)??

Clarify if i am wrong.

On May 30, 9:52 am, Aakash Johari <[email protected]> wrote:
> At the each level, traversed by BFS, you will have to check whether the
> vertex in this level has the element same as it found in the previous level.
> If it is different, then count it.
>
> On Sun, May 29, 2011 at 10:43 PM, anshu mishra 
> <[email protected]>wrote:
>
>
>
>
>
>
>
>
>
> > @piyush
>
> > void bfs(int mat[][n], bool flag[][n], int i, int j)
> > {
> > queue.push(mat[i][j]);
> > while (!q.empty())
> > {
> > x = q.top();
> > q.pop();
> > add top bottom, left right element in qeuue if their flag is true and their
> > value is equal to x and mark their flag false;
> > }
> > }
>
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> --
> -Aakash Johari
> (IIIT Allahabad)

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