LeetCode 73 - Set Matrix Zeroes
LeetCode 73 - Set Matrix Zeroes
Table of Contents
Problem Statement
Given an m x n integer matrix matrix, if an element is 0, set its entire row and column to 0’s.
You must do it in place.
Using In-place
Explanation
- Use First Row and First Column as Markers:
- Traverse the matrix, and if an element is
0, mark the corresponding row and column in the first row and first column.
- Traverse the matrix, and if an element is
- Set the Matrix Elements to Zero:
- Traverse the matrix again, and based on the markers, set the elements in the corresponding rows and columns to
0.
- Traverse the matrix again, and based on the markers, set the elements in the corresponding rows and columns to
- Handle the First Row and First Column Separately:
- Since they are used as markers, use separate flags to determine if they themselves need to be set to
0.
- Since they are used as markers, use separate flags to determine if they themselves need to be set to
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class Solution {
public:
void setZeroes(vector<vector<int>>& matrix) {
int m = matrix.size(), n = matrix[0].size();
bool firstRowZero = false, firstColZero = false;
for (int i = 0; i < m; i++)
if (matrix[i][0] == 0) firstColZero = true;
for (int j = 0; j < n; j++)
if (matrix[0][j] == 0) firstRowZero = true;
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (matrix[i][j] == 0) {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
for (int i = 1; i < m; i++) {
for (int j = 1; j < n; j++) {
if (matrix[i][0] == 0 || matrix[0][j] == 0)
matrix[i][j] = 0;
}
}
if (firstRowZero)
fill(matrix[0].begin(), matrix[0].end(), 0);
if (firstColZero)
for (int i = 0; i < m; i++) matrix[i][0] = 0;
}
};
Conclusion
Time Complexity: 𝑂(𝑚 × 𝑛) — The matrix is traversed multiple times, but the traversal remains linear relative to its size.
Space Complexity: 𝑂(1) — No additional space is used beyond a few extra variables.