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81. Top K Frequent Elements

mediumAsked at Reddit

Return the k most frequent elements in an array. Reddit uses this to test count + heap design — the exact shape of their trending-words detector in subreddit titles over a time window.

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

Source citations

Public interview reports confirming this problem appears in Reddit loops.

  • Glassdoor (2026-Q1)— Reddit ranking-team on-site favorite.
  • Blind (2025-12)— Reported as Reddit trending-detection canonical.

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
  • -10^4 <= nums[i] <= 10^4
  • 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 entries by count

Count occurrences; sort entries by count descending; return top k.

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

Tradeoff: O(m log m) sort. m = unique count.

2. Bucket sort by frequency (optimal)

Count; create n+1 buckets indexed by frequency. Walk buckets high to low; collect first k.

Time
O(n)
Space
O(n)
function topKFrequent(nums, k) {
  const count = new Map();
  for (const n of nums) count.set(n, (count.get(n) || 0) + 1);
  const buckets = Array.from({length: nums.length + 1}, () => []);
  for (const [val, c] of count) buckets[c].push(val);
  const out = [];
  for (let i = buckets.length - 1; i >= 0 && out.length < k; i--) {
    for (const v of buckets[i]) {
      if (out.length < k) out.push(v);
    }
  }
  return out;
}

Tradeoff: Linear. Bucket sort exploits the bounded frequency range (0..n).

Reddit-specific tips

Reddit interviewers will accept any version but reward the bucket-sort O(n). Bonus signal: connect to their trending-keywords pipeline where bucket sort beats heap when n is large but k is moderate.

Common mistakes

  • Allocating n buckets but indexing into a count up to n inclusive (off by one).
  • Adding all bucket contents (must respect k).
  • Using heap of size n (O(n log n)); use heap of size k for O(n log k).

Follow-up questions

An interviewer at Reddit may pivot to one of these next:

  • Sort characters by frequency (LC 451).
  • Top K Frequent Words (LC 692).
  • Kth largest element (LC 215).

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Output

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FAQ

Heap vs. bucket sort?

Heap O(n log k); bucket O(n). Bucket wins when frequencies have a known bounded range (which is always n here).

What if multiple values tie for k-th?

Problem guarantees unique answer, so this can't happen.

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