How do you solve Sum of Elements With Frequency Divisible by K on LeetCode?

Sum of Elements With Frequency Divisible by K (LeetCode #3712) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a hash map allows efficient frequency counting.

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

The optimal solution for Sum of Elements With Frequency Divisible by K runs in O(n) time and uses O(n) space. Counting frequencies in one pass and summing in another leads to O(n) time complexity.

What is the brute force solution for Sum of Elements With Frequency Divisible by K?

The brute force approach for Sum of Elements With Frequency Divisible by K is Brute Force, with O(n²) time complexity and O(1) space complexity. Count the frequency of each element and check if it's divisible by k. If it is, add it to the sum. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Sum of Elements With Frequency Divisible by K?

Sum of Elements With Frequency Divisible by K uses the following concepts: Array, Hash Table, Counting. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Sum of Elements With Frequency Divisible by K?

Explain your thought process clearly. Consider edge cases like empty arrays. Discuss time and space complexities.

What are common mistakes when solving Sum of Elements With Frequency Divisible by K?

Forgetting to check divisibility before adding to the sum. Not using a hash map for frequency counting.

#3712

Sum of Elements With Frequency Divisible by K

Easy

ArrayHash TableCountingHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

Use a hash map to count frequencies in one pass, then sum elements based on their frequencies in a second pass.

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Algorithm

3 steps

1Step 1: Create a frequency map of elements in nums.
2Step 2: Iterate through the frequency map and check if each frequency is divisible by k.
3Step 3: If divisible, add the element multiplied by its frequency to the total sum.

solution.py4 lines

1def sumDivisibleByK(nums, k):
2    from collections import Counter
3    freq_map = Counter(nums)
4    return sum(num * freq for num, freq in freq_map.items() if freq % k == 0)

ℹ

Complexity note: Counting frequencies in one pass and summing in another leads to O(n) time complexity.

1Using a hash map allows efficient frequency counting.
2Divisibility checks can be done in a single pass after counting.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.