Algorithms click when state becomes visible

The first structure answers “how often does this value appear?” The second structure reverses that relationship and answers “which values appeared this many times?” That reversal is the useful step. Once frequency becomes an array index, the most frequent values live near the end, so retrieving the top values is a backward walk rather than a full comparison sort.

Why are there n + 1 buckets?

A value cannot appear more than the length of the input. For an array of six items, valid frequencies are 1 through 6. Index 0 stays unused, which makes the code easier to read because bucket index 3 literally means “appeared three times.” We distribute every distinct value once, then visit buckets from high frequency to low frequency until the result contains k values.

function topKFrequent(nums: number[], k: number): number[] {
  const counts = new Map<number, number>();
  for (const number of nums) {
    counts.set(number, (counts.get(number) ?? 0) + 1);
  }

  const buckets: number[][] = Array.from(
    { length: nums.length + 1 },
    () => [],
  );
  for (const [number, count] of counts) buckets[count].push(number);

  const result: number[] = [];
  for (let count = buckets.length - 1; count >= 0; count--) {
    for (const number of buckets[count]) {
      result.push(number);
      if (result.length === k) return result;
    }
  }
  return result;
}