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18. Top K Frequent Elements

mediumAsked at LINE

Return the k most frequent elements from an array — LINE uses this to gauge whether you reach for bucket sort by frequency, the same shape behind top-k sticker ranking.

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

Problem

Given an integer array nums and an integer k, return the k most frequent elements in any order. Your solution must run in better than O(n log n) time.

Constraints

  • 1 <= nums.length <= 10^5
  • k is in the range [1, number of unique elements]
  • 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 frequencies in a map, sort the unique keys by descending frequency, take the first k.

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

Tradeoff:

2. Bucket sort by frequency

Count frequencies, then index buckets[freq] = [keys]. Scan buckets from high frequency down and collect until you have k elements. Linear in n because frequency is bounded by n.

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 [num, f] of count) buckets[f].push(num);
  const out = [];
  for (let f = buckets.length - 1; f >= 0 && out.length < k; f--) {
    for (const n of buckets[f]) {
      out.push(n);
      if (out.length === k) break;
    }
  }
  return out;
}

Tradeoff:

LINE-specific tips

At LINE, mention that ranking top-k stickers per chat room by usage frequency is the production version of this exact pipeline — sticker-delivery framing wins.

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Output

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