LeetCode-in-Java

3712. Sum of Elements With Frequency Divisible by K

Easy

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.

The frequency of an element x is the number of times it occurs in the array.

Example 1:

Input: nums = [1,2,2,3,3,3,3,4], k = 2

Output: 16

Explanation:

• The number 1 appears once (odd frequency).
• The number 2 appears twice (even frequency).
• The number 3 appears four times (even frequency).
• The number 4 appears once (odd frequency).

So, the total sum is 2 + 2 + 3 + 3 + 3 + 3 = 16.

Example 2:

Input: nums = [1,2,3,4,5], k = 2

Output: 0

Explanation:

There are no elements that appear an even number of times, so the total sum is 0.

Example 3:

Input: nums = [4,4,4,1,2,3], k = 3

Output: 12

Explanation:

• The number 1 appears once.
• The number 2 appears once.
• The number 3 appears once.
• The number 4 appears three times.

So, the total sum is 4 + 4 + 4 = 12.

Constraints:

• 1 <= nums.length <= 100
• 1 <= nums[i] <= 100
• 1 <= k <= 100

Solution

public class Solution {
    public int sumDivisibleByK(int[] nums, int k) {
        int max = 0;
        int sum = 0;
        for (int num : nums) {
            max = Math.max(num, max);
        }
        int[] cnt = new int[max + 1];
        for (int num : nums) {
            cnt[num]++;
        }
        for (int i = 1; i < cnt.length; i++) {
            if (cnt[i] != 0 && cnt[i] % k == 0) {
                sum += i * cnt[i];
            }
        }
        return sum;
    }
}

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