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21. Top K Frequent Elements

mediumAsked at Indeed

Return the k most frequent elements — a core primitive in Indeed's trending search query and popular job-title surfacing pipelines.

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]
  • 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 with a hash map then sort entries by count 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(e=>e[0]);
}

Tradeoff:

2. Bucket sort (O(n))

Place elements into frequency buckets indexed 0..n; scan from the highest bucket down to collect k elements without any comparison 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 [num, cnt] of freq) buckets[cnt].push(num);
  const res = [];
  for (let i = buckets.length-1; i >= 0 && res.length < k; i--) {
    for (const n of buckets[i]) { res.push(n); if (res.length===k) break; }
  }
  return res;
}

Tradeoff:

Indeed-specific tips

Indeed expects bucket sort for this problem; relate it to their real-time trending-search counter that must produce top-k results in a single pass over impression logs.

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Output

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