23. Top K Frequent Elements
mediumAsked at SquareSurface the K most common payment failure codes from a transaction log — Square's Payments Risk team runs this query at dashboard refresh time to triage which error categories need immediate incident response.
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. Your algorithm must be better than O(n log n) time complexity.
Constraints
1 <= nums.length <= 10^5-10^4 <= nums[i] <= 10^4k is in the range [1, 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]Example 2
nums = [1], k = 1[1]Approaches
1. Hash map + sort
Count frequencies with a map, sort by frequency descending, take first k. O(n log n) — violates the problem's complexity requirement.
- 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(e => e[0]);
}Tradeoff:
2. Bucket sort (O(n))
Count frequencies; create buckets indexed by frequency (bucket[f] = list of nums with frequency f). Iterate buckets from high to low and collect k elements. O(n) time and space — meets the problem's requirement.
- 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 = new Array(nums.length + 1).fill(null).map(() => []);
for (const [num, f] of freq) buckets[f].push(num);
const res = [];
for (let i = buckets.length - 1; i >= 0 && res.length < k; i--) {
res.push(...buckets[i]);
}
return res.slice(0, k);
}Tradeoff:
Square-specific tips
The O(n log n) sort solution is the trap — the problem explicitly bans it. Square uses this to see whether you push toward bucket sort on your own or wait to be led. Name the approach before coding it: 'because frequency is bounded by array length, I can use the frequency as an array index.' That single sentence demonstrates the key insight clearly.
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