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19. Top K Frequent Elements

easyAsked at Etsy

Return the k most frequent elements — the direct model behind Etsy's 'best-selling items in a category' ranking that drives the search results page.

By Alex Chen, Founder, InterviewChamp.AI · Last verified

Problem

Given an integer array nums and an integer k, return the k most frequent elements. You may return the answer in any order.

Constraints

  • 1 <= nums.length <= 10^5
  • k is in the range [1, the number of unique elements in the array]
  • It is guaranteed that the answer is unique

Examples

Example 1

Input
nums = [1,1,1,2,2,3], k = 2
Output
[1,2]

Example 2

Input
nums = [1], k = 1
Output
[1]

Approaches

1. Brute force sort

Count frequencies with a map, convert to array of [value, count] pairs, sort descending by count, slice the first k. O(n log n) due to the sort.

Time
O(n log n)
Space
O(n)
function topKFrequent(nums, k) {
  const freq = new Map();
  for (const n of nums) freq.set(n, (freq.get(n) || 0) + 1);
  return [...freq.entries()]
    .sort((a, b) => b[1] - a[1])
    .slice(0, k)
    .map(e => e[0]);
}

Tradeoff:

2. Bucket sort (linear)

After building the frequency map, create n+1 buckets indexed by frequency. Place each unique element in its frequency bucket. Walk buckets from high to low, collecting elements until k are gathered. O(n) time.

Time
O(n)
Space
O(n)
function topKFrequent(nums, k) {
  const freq = new Map();
  for (const n of nums) freq.set(n, (freq.get(n) || 0) + 1);

  const buckets = Array.from({ length: nums.length + 1 }, () => []);
  for (const [val, cnt] of freq) buckets[cnt].push(val);

  const result = [];
  for (let i = buckets.length - 1; i >= 0 && result.length < k; i--) {
    result.push(...buckets[i]);
  }
  return result.slice(0, k);
}

Tradeoff:

Etsy-specific tips

Etsy uses this pattern for real-time trending categories. Mention the bucket-sort trick explicitly — it shows you recognize when sorting is unnecessary because the domain (frequency 1..n) gives you a natural index. The interviewer will often ask you to stream updates; be ready to switch to a min-heap approach for that variant.

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Output

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