Start BFS from a position which is 'o'. the neighbour elements are as
defined in the question. Mark neighbour as 'Z' if it is not a boundary and
either 'x' or 'o', else *failure*. If there is no more for the considered
connected component *and if it is* *success*, then mark all 'Z's to 'x'. Do
this
sorry.. pls ignore the above post.. that doesn't work..
On Wed, Jun 19, 2013 at 10:5
5 PM, bharat b wrote:
> If I understood the problem, the following works fine.
> if(A[i][j] == 'o' and it is not and edge element) {
> if(A[i][j] is surrounded by either 'x' or 'o') {
> A[i][j] = 'x';
> }
If I understood the problem, the following works fine.
if(A[i][j] == 'o' and it is not and edge element) {
if(A[i][j] is surrounded by either 'x' or 'o') {
A[i][j] = 'x';
}
}
On Mon, Jun 10, 2013 at 8:38 PM, Jai Shri Ram wrote:
> Given a 2D board containing 'X' and 'O', capture all regio
