Count indices with opposite parity

81.6% acceptance

May 13, 2026

You are given an integer array nums of length n. The score of an index i is defined as the number of indices j such that: i < j < n, and nums[i] and nums[j] have different parity (one is even and the other is odd). Return an integer array answer of length n, where answer[i] is the score of index i.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn count_opposite_parity(nums: Vec<i32>) -> Vec<i32> {
    let n = nums.len();
    let mut suffix_even = 0i32;
    let mut suffix_odd = 0i32;
    let mut ans = vec![0i32; n];
    for i in (0..n).rev() {
      ans[i] = if nums[i] % 2 == 0 { suffix_odd } else { suffix_even };
      if nums[i] % 2 == 0 { suffix_even += 1; } else { suffix_odd += 1; }
    }
    ans
  }
}