Given n
and m
which are the dimensions of a matrix initialized by zeros and given an array indices
where indices[i] = [ri, ci]
. For each pair of [ri, ci]
you have to increment all cells in row ri
and column ci
by 1.
Return the number of cells with odd values in the matrix after applying the increment to all indices
.
Input: n = 2, m = 3, indices = [[0,1],[1,1]] Output: 6 Explanation: Initial matrix = [[0,0,0],[0,0,0]]. After applying first increment it becomes [[1,2,1],[0,1,0]]. The final matrix will be [[1,3,1],[1,3,1]] which contains 6 odd numbers.
Input: n = 2, m = 2, indices = [[1,1],[0,0]] Output: 0 Explanation: Final matrix = [[2,2],[2,2]]. There is no odd number in the final matrix.
1 <= n <= 50
1 <= m <= 50
1 <= indices.length <= 100
0 <= indices[i][0] < n
0 <= indices[i][1] < m
impl Solution {
pub fn odd_cells(n: i32, m: i32, indices: Vec<Vec<i32>>) -> i32 {
let mut matrix = vec![vec![false; m as usize]; n as usize];
for index in indices {
let ri = index[0] as usize;
let ci = index[1] as usize;
for c in 0..(m as usize) {
matrix[ri][c] = !matrix[ri][c];
}
for r in 0..(n as usize) {
matrix[r][ci] = !matrix[r][ci];
}
}
matrix.iter().map(|r| r.iter().filter(|&&c| c).count() as i32).sum()
}
}