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19. Top K Frequent Elements

mediumAsked at Byju's

Return the k most frequent values in an integer array.

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. The algorithm's time complexity must be better than O(n log n) where n is the array's size.

Constraints

  • 1 <= nums.length <= 10^5
  • k is in the range [1, number of unique elements]

Examples

Example 1

Input
nums = [1,1,1,2,2,3], k = 2
Output
[1,2]

Example 2

Input
nums = [1], k = 1
Output
[1]

Approaches

1. Sort by frequency

Count with a map, then sort entries by descending frequency.

Time
O(n log n)
Space
O(n)
const m=new Map();
for(const n of nums) m.set(n,(m.get(n)||0)+1);
return [...m.entries()].sort((a,b)=>b[1]-a[1]).slice(0,k).map(e=>e[0]);

Tradeoff:

2. Bucket sort by count

Count frequencies, then bucket values by their count index. Walk buckets from highest to lowest and pull k values out.

Time
O(n)
Space
O(n)
function topKFrequent(nums, k) {
  const count = new Map();
  for (const n of nums) count.set(n, (count.get(n) || 0) + 1);
  const buckets = Array.from({length: nums.length + 1}, () => []);
  for (const [num, c] of count) buckets[c].push(num);
  const res = [];
  for (let i = buckets.length - 1; i >= 0 && res.length < k; i--)
    for (const num of buckets[i]) {
      res.push(num);
      if (res.length === k) break;
    }
  return res;
}

Tradeoff:

Byju's-specific tips

Byju's recommendation team uses top-k frequency aggregation for the 'most-watched lesson' carousel, so cite that during the design walk-through.

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