#1207 - Unique Number of Occurrences

EasyHash Map / Set

Given an array of integers nums, return true if the number of occurrences of each value in the array is unique, or false otherwise.

Input	Output
nums = [1,2,2,1,1,3]	true
nums = [1,2]	false
nums = [-3,0,1,-3,1,1,1,-3,10,0]	true

Solution Space

Try it

/**
 * Returns true if every value's occurrence count is unique.
 *
 * @param nums Array of integers (1 ≤ length ≤ 1000, −1000 ≤ nums[i] ≤ 1000)
 * @returns Whether all frequency counts are distinct
 */
export function uniqueOccurrences(nums: number[]): boolean {
  const freq = new Map<number, number>();

  for (const n of nums) {
    freq.set(n, (freq.get(n) ?? 0) + 1);
  }

  const counts = new Set(freq.values());

  // If a count appeared more than once, the Set will be smaller than the Map
  return counts.size === freq.size;
}

Approach: frequency map + set – count occurrences with a Map, then put all counts into a Set; if the Set and Map have the same size, every count is unique.

Time: O(n) – one pass to build the frequency map, one pass over its values.

Space: O(n) – worst-case every element is distinct, so the map has n entries.

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