Is Sum of Elements With Frequency Divisible by K easy or hard?

This problem is considered Easy. It focuses on recognizing the frequency counting pattern and using a hash map efficiently. Once the counts are computed, the divisibility check is straightforward.

Sum of Elements With Frequency Divisible by K Python/Java solution

Both Python and Java implementations use a dictionary or HashMap to count frequencies. After building the map, iterate through the key-value pairs and sum keys whose frequency is divisible by k. The logic remains identical across Python, Java, C++, Go, and TypeScript.

How to solve Sum of Elements With Frequency Divisible by K in O(n)?

Use a frequency hash map. Iterate through the array once to count occurrences of each number. After building the map, iterate over its entries and check whether the stored frequency is divisible by k. Add those elements to the sum. Each step uses constant-time hash operations, giving overall O(n) complexity.

What is the best approach for Sum of Elements With Frequency Divisible by K?

The best approach uses a hash map to count element frequencies. First iterate through the array and store counts in a map. Then check each element's frequency and add the element to the result if frequency % k == 0. This method runs in O(n) time with O(n) space.

Is Sum of Elements With Frequency Divisible by K asked at Google/Amazon/Meta?

Problems based on frequency counting and hash maps appear frequently in interviews at companies like Amazon, Google, and Meta. While this exact problem may vary, the underlying pattern of counting occurrences and applying a condition is a common interview theme.

What data structure is used in Sum of Elements With Frequency Divisible by K?

A hash table (hash map or dictionary) is the main data structure. It maps each element to its frequency in the array. This structure enables constant-time average updates and lookups, making the overall algorithm linear time.

What is the time complexity of Sum of Elements With Frequency Divisible by K?

The optimal solution runs in O(n) time. One pass builds the frequency map and another pass checks the divisibility condition for each unique element. Space complexity is O(n) because the hash map stores counts for distinct values.

3712. Sum of Elements With Frequency Divisible by K - Practice Online

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