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21. Top K Frequent Elements

easyAsked at Tripadvisor

Return the k most frequent values in an array — Tripadvisor applies this exact heap/bucket pattern to surface the top-k trending destinations or most-reviewed attractions on their homepage.

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. The answer is guaranteed to be unique.

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. Sort by frequency

Count frequencies with a hash map, then sort the unique elements by frequency descending, and take the first k.

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

Tradeoff:

2. Bucket sort (optimal)

Build frequency buckets indexed by count (max index = n). Iterate buckets from highest to lowest, collecting elements until k are gathered. Beats O(n log n) sort.

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, count] of freq) buckets[count].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:

Tripadvisor-specific tips

Tripadvisor's recommendation team cares deeply about ranking — this problem is a direct proxy for how you'd rank destinations by review count or booking frequency. Interviewers expect you to land on the bucket-sort O(n) solution or at least articulate why it beats a heap approach in this case. Bonus points if you mention that in streaming analytics (real-time trending destinations), a min-heap of size k is often more practical than bucket sort.

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