Delayed count of equal elements

Array Hash Table Counting

80.6% acceptance

Apr 3, 2026

You are given an integer array nums of length n and an integer k. For each index i, define the delayed count as the number of indices j such that: i + k < j <= n - 1, and nums[j] == nums[i] Return an array ans where ans[i] is the delayed count of index i.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn delayed_count(nums: Vec<i32>, k: i32) -> Vec<i32> {
    let n = nums.len();
    let gap = k as usize;
    let mut freq = vec![0i32; 100_001];
    let mut answer = vec![0; n];

    for index in (0..n).rev() {
      let delayed_index = index + gap + 1;
      if delayed_index < n {
        freq[nums[delayed_index] as usize] += 1;
      }
      answer[index] = freq[nums[index] as usize];
    }

    answer
  }
}