13. Set Matrix Zeroes.cpp

/*
Set Matrix Zeroes
=================

Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0's, and return the matrix.

You must do it in place.

Example 1:
Input: matrix = [[1,1,1],[1,0,1],[1,1,1]]
Output: [[1,0,1],[0,0,0],[1,0,1]]

Example 2:
Input: matrix = [[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]]
 
Constraints:
m == matrix.length
n == matrix[0].length
1 <= m, n <= 200
-231 <= matrix[i][j] <= 231 - 1
 
Follow up:
A straightforward 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>>& A) {
        int row = 1, col = 1;
        int n = A.size(), m = A[0].size();
        
        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < m; ++j) {
                if(A[i][j] == 0) {
                    
                    if(i == 0) row = 0;
                    if(j == 0) col = 0;
                    
                    if(i != 0 || j != 0) {
                        if(j != 0) A[0][j] = 0;
                        if(i != 0) A[i][0] = 0;
                    }
                    
                }
            }
        }
        
        for(int i = 1; i < n; ++i) {
            if(A[i][0] == 0) {
                for(int j = 0; j < m; ++j) {
                    A[i][j] = 0;
                }
            }
        }
        
        for(int j = 1; j < m; ++j) {
            if(A[0][j] == 0) {
                for(int i = 0; i < n; ++i) {
                    A[i][j] = 0;
                }
            }
        }
        
        if(row == 0) {
            for(int j = 0; j < m; ++j) {
                A[0][j] = 0;
            }
        }
        
        if(col == 0) {
            for(int i = 0; i < n; ++i) {
                A[i][0] = 0;
            }
        }
    }
};