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1207. Unique Number of Occurrences

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 $arr$.

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

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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();
    }
}

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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;
    }
};

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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
}

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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;
}

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