1207. Unique Number of Occurrences

leetcode easy

#array#csharp#easy#hash-table#leetcode

Task

Дан массив целых чисел arr. Верните true, если количество вхождений каждого значения в массиве уникально, или false в противном случае.

Пример:

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.

C# solution

matched/original

using System;
using System.Collections.Generic;
public class Solution {
    public bool UniqueOccurrences(int[] arr) {
        Dictionary<int, int> freq = new Dictionary<int, int>();
        foreach (int num in arr) {
            if (freq.ContainsKey(num)) {
                freq[num]++;
            } else {
                freq[num] = 1;
            }
        }
        HashSet<int> freqSet = new HashSet<int>(freq.Values);
        return freq.Count == freqSet.Count;
    }
}

C++ solution

auto-draft, review before submit

#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public bool UniqueOccurrences(vector<int>& arr) {
        unordered_map<int, int> freq = new unordered_map<int, int>();
        foreach (int num in arr) {
            if (freq.count(num)) {
                freq[num]++;
            } else {
                freq[num] = 1;
            }
        }
        HashSet<int> freqSet = new HashSet<int>(freq.Values);
        return freq.size() == freqSet.size();
    }
}

Java solution

matched/original

class Solution {
    public boolean uniqueOccurrences(int[] arr) {
        Map<Integer, Integer> freq = new HashMap<>();
        for (int num : arr) {
            freq.put(num, freq.getOrDefault(num, 0) + 1);
        }
        
        Set<Integer> freqSet = new HashSet<>(freq.values());
        
        return freq.size() == freqSet.size();
    }
}

JavaScript solution

matched/original

var uniqueOccurrences = function(arr) {
    let freq = new Map();
    for (let num of arr) {
        freq.set(num, (freq.get(num) || 0) + 1);
    }
    let freqSet = new Set(freq.values());
    return freq.size === freqSet.size;
};

Python solution

matched/original

class Solution:
    def uniqueOccurrences(self, arr: List[int]) -> bool:
        freq = {}
        for num in arr:
            freq[num] = freq.get(num, 0) + 1
        return len(freq) == len(set(freq.values()))

Go solution

matched/original

func uniqueOccurrences(arr []int) bool {
    freq := make(map[int]int)
    for _, num := range arr {
        freq[num]++
    }
    freqSet := make(map[int]struct{})
    for _, count := range freq {
        freqSet[count] = struct{}{}
    }
    return len(freq) == len(freqSet)
}

Explanation

Algorithm

1⃣Сохраните частоты элементов массива arr в хэш-таблице freq.

2⃣Итерируйтесь по хэш-таблице freq и вставьте частоты всех уникальных элементов массива arr в хэш-набор freqSet.

3⃣Верните true, если размер хэш-набора freqSet равен размеру хэш-таблицы freq, иначе верните false.

😎