17. Top K Frequent Elements
mediumAsked at ConfluentReturn 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.
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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 <= 10^5k is in the range [1, number of unique elements]Answer is guaranteed unique
Examples
Example 1
nums=[1,1,1,2,2,3], k=2[1,2]Example 2
nums=[1], k=1[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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