@rajnish:
i think u missed the point , the matrix is binary.. so how will you store -1
in it.
@pacific:
Will you solution work fine if 0 th row has all 1's and 0th column has
atleast one zero.??
On Mon, Mar 7, 2011 at 1:09 AM, yogesh kumar <yoku2...@gmail.com> wrote:
> @pacific: Good Algorithm
>
> // This is the preface solution of this problem [:)]
> // Time complexity: O(n^2)
> // Space complexity: O(1)
>
> public class BinaryMatrix
> {
> public static void main(String arg[])
> {
> int[][] matrix = new int[][]
> {{1,0,1,1,0},{0,1,1,1,0},{1,1,1,1,1},
> {1,0,1,1,1},{1,1,1,1,1}};
> int temp = 0;
> temp = matrix[0][0];
> for(int i=1; i<matrix.length;i++)
> {
> matrix[0][0] &= matrix[0][i];
> temp &= matrix[i][0];
> }
> for(int i=1;i<matrix.length;i++)
> {
> for(int j=1;j<matrix.length;j++)
> {
> matrix[0][i] &= matrix[j][i];
> matrix[i][0] &= matrix[i][j];
> }
> }
> for(int i=1;i<matrix.length;i++)
> {
> for(int j=1;j<matrix.length;j++)
> {
> matrix[i][j] = matrix[i][0] & matrix[0][j];
> }
> }
> for(int i=1; i<matrix.length;i++)
> {
> if(matrix[0][0] == 0)
> {
> matrix[0][i] = 0;
> }
> if(temp == 0)
> {
> matrix[i][0] = 0;
> }
> }
> matrix[0][0] &= temp;
> for(int i=0;i<matrix.length;i++)
> {
> for(int j=0;j<matrix.length;j++)
> {
> System.out.print(matrix[i][j] + " ");
> }
> System.out.print("\n");
> }
> }
> }
>
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