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19. Top K Frequent Elements

easyAsked at Spotify

Given an integer array, return the k most frequent elements — a canonical heap pattern that maps directly to how Spotify surfaces the top-K tracks on a Weekly Charts leaderboard.

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, number of unique elements]
  • The answer is guaranteed to be 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. Sort by frequency

Count frequencies with a hash map, sort entries descending by count, return the first k keys.

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(([num]) => num);
}

Tradeoff:

2. Bucket sort (optimal)

Build frequency buckets indexed by count; scan from highest bucket down to collect k elements in O(n) without sorting.

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

Tradeoff:

Spotify-specific tips

Spotify engineers ask this to probe whether you know when sorting is unnecessary. The bucket-sort insight — that count can't exceed array length, so you can index directly — earns the most credit. Tie your explanation to chart ranking: 'every stream event increments a counter; we want the top-K at query time without a full re-sort.'

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Output

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