Set Matrix Zeroes

We basically are required to iterate through the matrix at least twice:

• the first time is necessary to identify all the matrix values that have been set equal to 0
  • you cannot change while iterating, as there is no way to differentiate between existing 0 values and newly-converted 0 values
• the second time is necessary to convert all the relevant matrix values to 0

Now, in the first iteration, I realised that if matrix[i][j] == 0, we could simply convert:

• the value present in the first column (of the same row), matrix[i][0]
• the value present in the first row (of the same column), matrix[0][j]

and then on our second iteration, we can check if either matrix[i][0] == 0 or matrix[0][j] == 0 to determine if matrix[i][j] should be converted to 0.

Unfortunately, there does exist the edge case of matrix[0][0] as values on the first row and first column will end up converting it when equal to 0. Therefore, we can initialize a boolean col0 to check whether we should convert column 0 to all 0s.

code

class Solution {
public:
    void setZeroes(vector<vector<int>>& matrix) {
        bool col0 = false;

        for (int i = 0; i < matrix.size(); ++i) {
            if (matrix[i][0] == 0) {
                col0 = true;
            }
            for (int j = 1; j < matrix[0].size(); ++j) {
                if (matrix[i][j] == 0) {
                    matrix[i][0] = matrix[0][j] = 0;
                }
            }
        }

        for (int i = matrix.size()-1; i >= 0; --i) {
            for (int j = matrix[0].size()-1; j >= 1; --j) {
                if (matrix[i][0] == 0 || matrix[0][j] == 0) {
                    matrix[i][j] = 0;
                }
            }
            if (col0) {
                matrix[i][0] = 0;
            }
        }
    }
};

complexity

• Time: O(n*m) where n*m are the dimensions of the matrix
• Space: O(1)

time taken

• 9 minutes exactly