Cells with odd values in a matrix

Array Math Simulation

79.7% acceptance

Feb 25, 2026

1335

1560

There is an m x n matrix that is initialized to all 0's. There is also a 2D array indices where each indices[i] = [ri, ci] represents a 0-indexed location to perform some increment operations on the matrix. For each location indices[i], do both of the following: Increment all the cells on row ri. Increment all the cells on column ci. Given m, n, and indices, return the number of odd-valued cells in the matrix after applying the increment to all locations in indices.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn odd_cells(m: i32, n: i32, indices: Vec<Vec<i32>>) -> i32 {
    let m = m as usize;
    let n = n as usize;
    let mut row_parity = vec![0usize; m];
    let mut col_parity = vec![0usize; n];

    for idx in &indices {
      row_parity[idx[0] as usize] += 1;
      col_parity[idx[1] as usize] += 1;
    }

    let odd_rows = row_parity.iter().filter(|&&r| r % 2 == 1).count();
    let odd_cols = col_parity.iter().filter(|&&c| c % 2 == 1).count();
    let even_rows = m - odd_rows;
    let even_cols = n - odd_cols;

    // Cell (r, c) is odd if row_parity[r] + col_parity[c] is odd
    // = (odd_row, even_col) or (even_row, odd_col)
    (odd_rows * even_cols + even_rows * odd_cols) as i32
  }
}