Demo preview: These solutions pass the automated test suite but have not been human-reviewed. Treat trade-offs and prose as draft.

Mediumhashing

Top K Frequent Elements

Problem (restated)

Return the k most frequent elements in any order.

Intuition

Count frequencies, then bucket by frequency so the densest buckets are at the end.

Approaches

Frequency buckets

Tested only

Time O(n)Space O(n)

Idea. Map value→count. buckets[c] holds values with count c. Scan buckets from high to low.

Walkthrough. [1,1,1,2,2,3], k=2 → 1 has freq 3, 2 has 2 → [1,2].

Trade-offs. Heap is O(n log k). Buckets hit average O(n) with integer frequencies.

Solution

export function topKFrequent(nums: number[], k: number): number[] {
  const freq = new Map<number, number>();
  for (const x of nums) freq.set(x, (freq.get(x) ?? 0) + 1);
  const buckets: number[][] = Array.from({ length: nums.length + 1 }, () => []);
  for (const [val, c] of freq) buckets[c]!.push(val);
  const out: number[] = [];
  for (let c = buckets.length - 1; c >= 0 && out.length < k; c--) {
    for (const v of buckets[c]!) {
      out.push(v);
      if (out.length === k) return out;
    }
  }
  return out;
}

export function topKFrequent(nums: number[], k: number): number[] {
  const freq = new Map<number, number>();
  for (const x of nums) freq.set(x, (freq.get(x) ?? 0) + 1);
  const buckets: number[][] = Array.from({ length: nums.length + 1 }, () => []);
  for (const [val, c] of freq) buckets[c]!.push(val);
  const out: number[] = [];
  for (let c = buckets.length - 1; c >= 0 && out.length < k; c--) {
    for (const v of buckets[c]!) {
      out.push(v);
      if (out.length === k) return out;
    }
  }
  return out;
}

Reflection

Which cue made you pick this pattern in under 90 seconds?
What input would break a wrong invariant?