Given a m x n matrix, if an element is 0, set its entire row and column to 0. Do it in place.
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?
void setZeroes(vector<vector<int>>& matrix) { int m = matrix.size(), n = matrix[0].size(); vector<bool> clearRow(m), clearCol(n); for(int i=0; i<m; i++) { for(int j=0; j<n; j++){ if(!matrix[i][j]) { clearRow[i] = true; clearCol[j] = true; } } } for(int i=0; i<m; i++) { for(int j=0; j<n; j++) { if(clearRow[i] || clearCol[j]) { matrix[i][j] = 0;; } } } }
Space Complexity: O(M+N).
void setZeroes(vector<vector<int>>& matrix) { int m = matrix.size(), n = matrix[0].size(); bool clear1stRow = false, clear1stCol = false; for(int i=0; i<m; i++) { if(!matrix[i][0]) { clear1stCol = true; break; } } for(int j=0; j<n; j++){ if(!matrix[0][j]) { clear1stRow = true; break; } } for(int i=1; i<m; i++) { for(int j=1; j<n; j++) { if(!matrix[i][j]) { matrix[0][j] = 0; matrix[i][0] = 0; } } } for(int i=1; i<m; i++) { for(int j=1; j<n; j++) { if(!matrix[i][0] || !matrix[0][j]) { matrix[i][j] = 0; } } } if(clear1stRow) { matrix[0].assign(n,0); } if(clear1stCol) { for(int i=0; i<m; i++) { matrix[i][0] = 0; } } }
Space Complexity: O(1)