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16. Top K Frequent Elements

mediumAsked at Glassdoor

Surfacing the highest-rated companies out of millions of reviews is Glassdoor's bread and butter — this heap-based top-K problem tests whether you can rank by frequency without sorting everything first.

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 nums]
  • The answer is guaranteed to be unique

Examples

Example 1

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

Explanation: 1 appears 3 times, 2 appears twice — both beat 3's single occurrence.

Example 2

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

Approaches

1. Sort by frequency

Count frequencies with a map, then sort entries by count descending and take 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(([num]) => num);
}

Tradeoff:

2. Bucket sort (linear)

Frequency can range from 1 to n, so use bucket indices. Place each element in the bucket at index equal to its count, then scan buckets right-to-left collecting k elements. 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 [num, count] of freq) buckets[count].push(num);

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

Tradeoff:

Glassdoor-specific tips

Glassdoor's ranking features depend on surfacing high-signal content quickly. Show that you know the bucket-sort trick drops the complexity to O(n) — interviewers there frequently push back with 'can you do better than n log n?' if you stop at the sort approach. Tie your explanation to how frequency caps at array length, which makes bounded buckets possible.

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Output

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