Sum of elements with frequency divisible by k

Array Hash Table Counting

77.6% acceptance

Feb 24, 2026

You are given an integer array nums and an integer k. Return an integer denoting the sum of all elements in nums whose frequency is divisible by k, or 0 if there are no such elements. Note: An element is included in the sum exactly as many times as it appears in the array if its total frequency is divisible by k.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

impl Solution {
  pub fn sum_divisible_by_k(nums: Vec<i32>, k: i32) -> i32 {
    let mut freq = std::collections::HashMap::new();
    for &x in &nums {
      *freq.entry(x).or_insert(0) += 1;
    }
    freq.iter()
      .filter(|&(_, cnt)| cnt % k == 0)
      .map(|(&val, &cnt)| val * cnt)
      .sum()
  }
}