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17. Top K Frequent Elements

mediumAsked at Confluent

Return the k most frequent elements — Confluent uses it because top-K aggregation is one of the bread-and-butter ksqlDB queries running over a Kafka topic.

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, number of unique elements]
  • Answer is guaranteed 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 occurrences, sort unique values by descending count, return 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 count

Count occurrences then place each unique value into a bucket indexed by its count. Sweep buckets from high to low and collect k values.

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

Tradeoff:

Confluent-specific tips

Confluent will push you toward the streaming variant — explain how each partition keeps its own top-K heap and a single coordinator merges them, so partition assignment changes don't collapse the global ranking.

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Output

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