19. Top K Frequent Elements
mediumAsked at ByteDanceReturn the k most frequent elements in an array — ByteDance uses it as a direct proxy for top-K trending content selection in TikTok feeds.
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, and the answer is guaranteed to be unique.
Constraints
1 <= nums.length <= 10^5k is in [1, number of distinct elements]
Examples
Example 1
nums = [1,1,1,2,2,3], k = 2[1,2]Example 2
nums = [1], k = 1[1]Approaches
1. Count, sort, take k
Build a frequency map, sort entries by count, slice the top 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(([v])=>v);Tradeoff:
2. Bucket sort by frequency
Build a frequency map, bucket values by frequency (1..n), then read buckets from highest down until k values are collected.
- Time
- O(n)
- Space
- O(n)
function topKFrequent(nums, k) {
const count = new Map();
for (const x of nums) count.set(x, (count.get(x) || 0) + 1);
const buckets = Array.from({ length: nums.length + 1 }, () => []);
for (const [val, c] of count) buckets[c].push(val);
const out = [];
for (let f = buckets.length - 1; f > 0 && out.length < k; f--) {
for (const v of buckets[f]) {
out.push(v);
if (out.length === k) break;
}
}
return out;
}Tradeoff:
ByteDance-specific tips
ByteDance interviewers love the bucket-sort framing because it mirrors how their trending-hashtag service partitions counts by frequency rather than full-sorting them.
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