17. Top K Frequent Elements
mediumAsked at CoinbaseSurface the K most-traded assets on an exchange — Coinbase uses this problem to test whether you reach for a heap over a naive sort when ranking trade volumes across thousands of instruments.
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, the number of unique elements in the array]The answer is guaranteed to be unique
Examples
Example 1
nums = [1,1,1,2,2,3], k = 2[1,2]Explanation: 1 appears 3 times, 2 appears 2 times — both top-2.
Example 2
nums = [1], k = 1[1]Approaches
1. Sort by frequency
Build a frequency map, collect entries, sort descending by count, return first k keys.
- Time
- O(n log n)
- Space
- O(n)
function topKFrequent(nums, k) {
const freq = new Map();
for (const n of nums) freq.set(n, (freq.get(n) || 0) + 1);
return [...freq.entries()]
.sort((a, b) => b[1] - a[1])
.slice(0, k)
.map(([num]) => num);
}Tradeoff:
2. Bucket sort (optimal)
Use an array of buckets indexed by frequency (max = n). Iterate buckets from high to low to collect k results in O(n).
- Time
- O(n)
- Space
- O(n)
function topKFrequent(nums, k) {
const freq = new Map();
for (const n of nums) freq.set(n, (freq.get(n) || 0) + 1);
const buckets = Array.from({ length: nums.length + 1 }, () => []);
for (const [num, count] of freq) buckets[count].push(num);
const result = [];
for (let i = buckets.length - 1; i >= 0 && result.length < k; i--) {
result.push(...buckets[i]);
}
return result.slice(0, k);
}Tradeoff:
Coinbase-specific tips
Coinbase interviewers pay close attention to whether you recognise the bucket-sort escape hatch. When the range of possible frequencies is bounded by n, you can beat the O(n log n) sort — a signal that you think about constraints the way trading-system engineers think about throughput ceilings.
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