73. Set Matrix Zeroes.cpp

Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in-place.

Example 1:

Input:
[
  [1,1,1],
  [1,0,1],
  [1,1,1]
]
Output:
[
  [1,0,1],
  [0,0,0],
  [1,0,1]
]
Example 2:

Input:
[
  [0,1,2,0],
  [3,4,5,2],
  [1,3,1,5]
]
Output:
[
  [0,0,0,0],
  [0,4,5,0],
  [0,3,1,0]
]
Follow up:

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?








class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        bool fr = false, fc = false;  // check if first row or column contains a zero
        
        for (int i = 0; i < m; i++) // col 0
            if (matrix[i][0] == 0) fc = true;
        
        for(int j = 0; j < n; j++)  // row 0
            if (matrix[0][j] == 0) fr = true;
        
        for (int i = 1; i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[i][j] == 0){
                    matrix[0][j] = 0;
                    matrix[i][0] = 0;
                }
            }
        }
        
        for (int i = 1;i < m; i++) {
            for (int j = 1; j < n; j++) {
                if (matrix[0][j] == 0 || matrix[i][0] == 0) matrix[i][j] = 0;
            }
        }
        
        if (fr){
            for (int j = 0; j < n; j++) matrix[0][j] = 0;
        }
        
        if (fc){
            for (int i = 0; i < m; i++) matrix[i][0] = 0;
        }
    }
};





/*
Brute force approach would be to create two temporary arrays for row and col and marking them on the basis of 0s and 1s
*/