10. Top K Frequent Elements
mediumAsked at AdyenGiven an array, return the k most frequent elements.
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 <= 10^5k is in the range [1, number of unique elements]
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 frequencies then sort entries descending.
- Time
- O(n log n)
- Space
- O(n)
const f = new Map();
for (const x of nums) f.set(x, (f.get(x) || 0) + 1);
return [...f.entries()].sort((a,b) => b[1] - a[1]).slice(0,k).map(e => e[0]);Tradeoff:
2. Bucket sort by frequency
Place each value in a bucket indexed by its frequency, then sweep from the top.
- Time
- O(n)
- Space
- O(n)
function topKFrequent(nums, k) {
const f = new Map();
for (const x of nums) f.set(x, (f.get(x) || 0) + 1);
const buckets = Array.from({length: nums.length + 1}, () => []);
for (const [v, c] of f) buckets[c].push(v);
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:
Adyen-specific tips
Adyen frames this as a top-merchants-by-volume problem — they prefer the bucket-sort variant because routing tables prioritize sub-linear scans over heap maintenance.
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