Class Solution
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 <= 1001 <= nums[i] <= 1001 <= k <= 100
-
Constructor Summary
Constructors -
Method Summary
-
Constructor Details
-
Solution
public Solution()
-
-
Method Details
-
sumDivisibleByK
public int sumDivisibleByK(int[] nums, int k)
-