Given a

m

x

n

matrix, if an element is 0, set its entire row and column to 0. Do it in place.

class Solution {
    public void setZeroes(int[][] matrix) {
        Set<Integer> rows = new HashSet<>();
        Set<Integer> cols = new HashSet<>();

        int n = matrix.length, m = matrix[0].length;
        for (int i = 0; i < n; i++){
            rows.add(i);
        }
        for (int j = 0; j < m; j++){
            cols.add(j);
        }

        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                if (matrix[i][j] == 0){
                    rows.remove(i);
                    cols.remove(j);
                }
            }
        }


        for (int i = 0; i < n; i++){
            for (int j = 0; j < m; j++){
                if (rows.contains(i) && cols.contains(j)) continue;
                matrix[i][j] = 0;
            }
        }


    }
}

Follow up:

Did you use extra space?
A straight forward solution using O(mn) space is probably a bad idea.
A simple improvement uses O(m+n) space, but still not the best solution.
Could you devise a constant space solution?

public class Solution {
public void setZeroes(int[][] matrix) {
    boolean fr = false,fc = false;
    for(int i = 0; i < matrix.length; i++) {
        for(int j = 0; j < matrix[0].length; j++) {
            if(matrix[i][j] == 0) {
                if(i == 0) fr = true;
                if(j == 0) fc = true;
                matrix[0][j] = 0;
                matrix[i][0] = 0;
            }
        }
    }
    for(int i = 1; i < matrix.length; i++) {
        for(int j = 1; j < matrix[0].length; j++) {
            if(matrix[i][0] == 0 || matrix[0][j] == 0) {
                matrix[i][j] = 0;
            }
        }
    }
    if(fr) {
        for(int j = 0; j < matrix[0].length; j++) {
            matrix[0][j] = 0;
        }
    }
    if(fc) {
        for(int i = 0; i < matrix.length; i++) {
            matrix[i][0] = 0;
        }
    }

}