Medium
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.
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:
O(mn)
space is probably a bad idea.O(m + n)
space, but still not the best solution.impl Solution {
pub fn set_zeroes(matrix: &mut Vec<Vec<i32>>) {
let m = matrix.len();
let n = matrix[0].len();
let mut row0 = false;
let mut col0 = false;
// Check if 0th column needs to be marked all 0s in future
for row in matrix.iter() {
if row[0] == 0 {
col0 = true;
break;
}
}
// Check if 0th row needs to be marked all 0s in future
for j in 0..n {
if matrix[0][j] == 0 {
row0 = true;
break;
}
}
// Store the signals in 0th row and column
for i in 1..m {
for j in 1..n {
if matrix[i][j] == 0 {
matrix[i][0] = 0;
matrix[0][j] = 0;
}
}
}
// Mark 0 for all cells based on signal from 0th row and 0th column
for i in 1..m {
for j in 1..n {
if matrix[i][0] == 0 || matrix[0][j] == 0 {
matrix[i][j] = 0;
}
}
}
// Set 0th column
if col0 {
for i in 0..m {
matrix[i][0] = 0;
}
}
// Set 0th row
if row0 {
for j in 0..n {
matrix[0][j] = 0;
}
}
}
}