Question

1207. Unique Number of Occurrences

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Description

Given an array of integers arr, return true _if the number of occurrences of each value in the array is unique or _false_ otherwise_.

 

Example 1:

Input: arr = [1,2,2,1,1,3]
Output: true
Explanation: The value 1 has 3 occurrences, 2 has 2 and 3 has 1. No two values have the same number of occurrences.

Example 2:

Input: arr = [1,2]
Output: false

Example 3:

Input: arr = [-3,0,1,-3,1,1,1,-3,10,0]
Output: true

 

Constraints:

• 1 <= arr.length <= 1000
• -1000 <= arr[i] <= 1000

Solutions

Solution 1: Hash Table

We use a hash table $cnt$ to count the frequency of each number in the array $arr$, and then use another hash table $vis$ to count the types of frequencies. Finally, we check whether the sizes of $cnt$ and $vis$ are equal.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $arr$.

class Solution:
  def uniqueOccurrences(self, arr: List[int]) -> bool:
    cnt = Counter(arr)
    return len(set(cnt.values())) == len(cnt)

class Solution {
  public boolean uniqueOccurrences(int[] arr) {
    Map<Integer, Integer> cnt = new HashMap<>();
    for (int x : arr) {
      cnt.merge(x, 1, Integer::sum);
    }
    return new HashSet<>(cnt.values()).size() == cnt.size();
  }
}

class Solution {
public:
  bool uniqueOccurrences(vector<int>& arr) {
    unordered_map<int, int> cnt;
    for (int& x : arr) {
      ++cnt[x];
    }
    unordered_set<int> vis;
    for (auto& [_, v] : cnt) {
      if (vis.count(v)) {
        return false;
      }
      vis.insert(v);
    }
    return true;
  }
};

func uniqueOccurrences(arr []int) bool {
  cnt := map[int]int{}
  for _, x := range arr {
    cnt[x]++
  }
  vis := map[int]bool{}
  for _, v := range cnt {
    if vis[v] {
      return false
    }
    vis[v] = true
  }
  return true
}

function uniqueOccurrences(arr: number[]): boolean {
  const cnt: Map<number, number> = new Map();
  for (const x of arr) {
    cnt.set(x, (cnt.get(x) || 0) + 1);
  }
  return cnt.size === new Set(cnt.values()).size;
}