Check whether the frequency of the elements in the Array is unique or not

Given an array arr[] of N integers, the task is to check whether the frequency of the elements in the array is unique or not, or in other words, there are no two distinct numbers in an array with equal frequency. If all the frequency is unique then Print “YES”, else Print “NO”.

Examples:

Input: N = 5, arr = [1, 1, 2, 5, 5]
Output: false
Explanation: The array contains 2 (1’s), 1 (2’s), and 2 (5’s) since the frequency of 1 and 5 are the same i.e. 2 times. Therefore, this array does not satisfy the condition.

Input: N = 10, arr = [2, 2, 5, 10, 1, 2, 10, 5, 10, 2]
Output: true
Explanation:

• Number of 1’s -> 1
• Number of 2’s -> 4
• Number of 5’s -> 2
• Number of 10’s -> 3.

Since the number of occurrences of elements present in the array is unique. Therefore, this array satisfies the condition.

Approach: To solve the problem follow the below idea:

Calculate the frequency of each element of the array and check whether all the frequencies are unique or not.

Follow the steps to solve the problem:

• Calculate the frequency of each element and store it in the map.
• Now we have to check the frequencies present in the map are unique or not.
• To do so we can insert all the frequencies in the set and compare the size of the set and map.
• As the set store only unique elements so if the size of the set and the size of the map is equal then our answer is yes else our answer is no because there is at least one frequency that is repeating.

Below is the implementation of the above approach:

C++

// C++ code for the above approach:
#include <bits/stdc++.h>
using namespace std;

// Function to check if all occurences of
// elements are unique or not
bool isFrequencyUnique(int n, int arr[])
{

    // map to store frequencies of elements
    // in an array
    unordered_map<int, int> mp;
    for (int i = 0; i < n; i++) {
        mp[arr[i]]++;
    }

    // set to store the unique frequencies
    unordered_set<int> st;
    for (auto x : mp) {
        st.insert(x.second);
    }

    // Returns true if unique frequencies is equal
    // to total frequencies i.e all frequencies
    // are unique
    return st.size() == mp.size();
}

// Drivers code
int main()
{
    int arr[] = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
    int n = sizeof(arr) / sizeof(arr[0]);

    // Function Call
    if (isFrequencyUnique(n, arr)) {
        cout << "YES\n";
    }
    else {
        cout << "NO\n";
    }
    return 0;
}

Java

import java.util.*;

public class GFG {

    // Function to check if all occurrences of elements are
    // unique or not
    static boolean isFrequencyUnique(int n, int[] arr)
    {
        // Map to store frequencies of elements in an array
        Map<Integer, Integer> mp = new HashMap<>();
        for (int i = 0; i < n; i++) {
            mp.put(arr[i], mp.getOrDefault(arr[i], 0) + 1);
        }

        // Set to store the unique frequencies
        Set<Integer> st = new HashSet<>();
        for (Map.Entry<Integer, Integer> entry :
             mp.entrySet()) {
            st.add(entry.getValue());
        }

        // Returns true if unique frequencies are equal to
        // total frequencies i.e., all frequencies are
        // unique
        return st.size() == mp.size();
    }

    // Driver code
    public static void main(String[] args)
    {
        int[] arr = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
        int n = arr.length;

        // Function Call
        if (isFrequencyUnique(n, arr)) {
            System.out.println("YES");
        }
        else {
            System.out.println("NO");
        }
    }
}

// This code is contributed by shivamgupta0987654321

Python

# Function to check if all occurrences of
# elements are unique or not
def isFrequencyUnique(arr):
    # Dictionary to store frequencies of elements
    freq_dict = {}

    # Count the frequencies of elements
    for num in arr:
        if num in freq_dict:
            freq_dict[num] += 1
        else:
            freq_dict[num] = 1

    # Set to store the unique frequencies
    unique_freqs = set(freq_dict.values())

    # Returns True if unique frequencies are equal
    # to the total frequencies, i.e., all frequencies
    # are unique
    return len(unique_freqs) == len(freq_dict)

# Driver code
arr = [2, 2, 5, 10, 1, 2, 10, 5, 10, 2]
# Function Call
if isFrequencyUnique(arr):
    print("YES")
else:
    print("NO")

C#

// C# code for the above approach:

using System;
using System.Collections.Generic;

class Program
{
    // Function to check if all occurences of
    // elements are unique or not
    static bool IsFrequencyUnique(int n, int[] arr)
    {
        // dictionary to store frequencies of elements
        // in an array
        Dictionary<int, int> mp = new Dictionary<int, int>();
        for (int i = 0; i < n; i++)
        {
            if (mp.ContainsKey(arr[i]))
            {
                mp[arr[i]]++;
            }
            else
            {
                mp[arr[i]] = 1;
            }
        }
        // set to store the unique frequencies
        HashSet<int> st = new HashSet<int>();
        foreach (var x in mp)
        {
            st.Add(x.Value);
        }
        // Returns true if unique frequencies is equal
        // to total frequencies i.e all frequencies
        // are unique
        return st.Count == mp.Count;
    }
    // Drivers code
    static void Main(string[] args)
    {
        int[] arr = { 2, 2, 5, 10, 1, 2, 10, 5, 10, 2 };
        int n = arr.Length;
        // Function Call
        if (IsFrequencyUnique(n, arr))
        {
            Console.WriteLine("YES");
        }
        else
        {
            Console.WriteLine("NO");
        }
    }
}

Javascript

function isFrequencyUnique(arr) {
    // Map to store frequencies of
    // elements in an array
    const frequencyMap = new Map();
    for (const num of arr) {
        if (!frequencyMap.has(num)) {
            frequencyMap.set(num, 0);
        }
        frequencyMap.set(num, frequencyMap.get(num) + 1);
    }
    // Set to store the unique frequencies
    const uniqueFrequencies = new Set();
    for (const freq of frequencyMap.values()) {
        uniqueFrequencies.add(freq);
    }
    return uniqueFrequencies.size === frequencyMap.size;
}
// Driver code
const arr = [2, 2, 5, 10, 1, 2, 10, 5, 10, 2];
// Function Call
if (isFrequencyUnique(arr)) {
    console.log("YES");
} else {
    console.log("NO");
}

Output

YES

Time Complexity: O(N), we just need to traverse the array and map for once only.
Auxillary space: O(N), we need a frequency map and set to check the frequency of elements is unique or not.