4068. Sum Of Elements With Frequency Divisible By K¶

4068. 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.

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¶

class Solution {
    public int sumDivisibleByK(int[] nums, int k) {
        int n = nums.length;
        int freq[] = new int[101];
        for (int ele : nums)
            freq[ele]++;
        int sum = 0;
        for (int i = 0; i <= 100; i++) 
            if (freq[i] % k == 0)
                sum += freq[i] * i;
        return sum;        
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Explanation¶

[Add detailed explanation here]