21. Top K Frequent Elements
mediumAsked at N26Given an integer array, return the k most frequent elements. N26 frames it as picking the top-k merchant categories per customer for category-spend insights.
By Alex Chen, Founder, InterviewChamp.AI · Last verified
Problem
Given an integer array nums and an integer k, return the k most frequent elements in any order. You must aim for better than O(n log n).
Constraints
1 <= nums.length <= 10^5k is in the range [1, number of unique elements]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. Count then sort
Hash-count, then sort all unique keys by frequency and take the top k.
- Time
- O(n log n)
- Space
- O(n)
const m = new Map();
for (const x of nums) m.set(x,(m.get(x)||0)+1);
return [...m.entries()]
.sort((a,b)=>b[1]-a[1])
.slice(0,k)
.map(([v])=>v);Tradeoff:
2. Bucket sort by frequency
Frequencies are bounded by n, so put each unique element into a bucket indexed by its count. Scan buckets from high to low until k items are collected.
- Time
- O(n)
- Space
- O(n)
function topKFrequent(nums, k) {
const m = new Map();
for (const x of nums) m.set(x, (m.get(x)||0)+1);
const buckets = Array.from({length: nums.length + 1}, () => []);
for (const [val, cnt] of m) buckets[cnt].push(val);
const ans = [];
for (let i = buckets.length - 1; i >= 0 && ans.length < k; i--) {
for (const v of buckets[i]) if (ans.length < k) ans.push(v);
}
return ans;
}Tradeoff:
N26-specific tips
N26 likes when you connect bucket sort to their spend-insights pipeline, where merchant-category counts are pre-aggregated by month and the dashboard only ever needs the top 5.
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