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22. Top K Frequent Elements

mediumAsked at Booking

Return the k most frequent elements from an array — Booking's search team applies this pattern to surface the top-k most-booked destinations or properties for a given city query.

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

Problem

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

Constraints

  • 1 <= nums.length <= 10^5
  • k is in the range [1, the number of unique elements in nums]
  • The answer is unique — there is only one set of k most frequent elements

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 unique elements by count descending, return 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 (linear)

Frequencies range [1..n]. Create n+1 buckets indexed by frequency; reverse-scan to collect top k. O(n) total.

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:

Booking-specific tips

Booking processes millions of search queries daily and the bucket-sort approach gets attention in systems design follow-ups — it avoids the log factor and scales with data. Be ready to explain why the bucket index equals frequency, not value.

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Output

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