Check if frequency of each element in given array is unique or not

Given an array arr[] of N positive integers where the integers are in the range from 1 to N, the task is to check whether the frequency of the elements in the array is unique or not. If all the frequency is unique then print “Yes”, else print “No”.

Examples:

Input: N = 5, arr[] = {1, 1, 2, 5, 5}
Output: No
Explanation: 
The array contains 2 (1’s), 1 (2’s) and 2 (5’s), since the number of 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: Yes
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 does not satisfy the condition.

 

Naive Approach: The idea is to check for every number from 1 to N whether it is present in the array or not. If yes, then count the frequency of that element in the array, and store the frequency in an array. At last, just check for any duplicate element in the array and print the output accordingly.



Time Complexity: O(N2)
Auxiliary Space: O(N) 

Efficient Approach: The idea is to use Hashing. Below are the steps:

  1. Traverse the given array arr[] and store the frequency of each element in a Map.
  2. Now traverse the map and check if the count of any element occurred more than once.
  3. If the count of any element in the above steps is more than one then print “No”, else print “Yes”.

Below is the implementation of the above approach:
 

C++

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// C++ code for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to check whether the
// frequency of elements in array
// is unique or not.
bool checkUniqueFrequency(int arr[],
                          int n)
{
 
    // Freq map will store the frequency
    // of each element of the array
    unordered_map<int, int> freq;
 
    // Store the frequency of each
    // element from the array
    for (int i = 0; i < n; i++) {
        freq[arr[i]]++;
    }
 
    unordered_set<int> uniqueFreq;
 
    // Check whether frequency of any
    // two or more elements are same
    // or not. If yes, return false
    for (auto& i : freq) {
        if (uniqueFreq.count(i.second))
            return false;
        else
            uniqueFreq.insert(i.second);
    }
 
    // Return true if each
    // frequency is unique
    return true;
}
 
// Driver Code
int main()
{
    // Given array arr[]
    int arr[] = { 1, 1, 2, 5, 5 };
    int n = sizeof arr / sizeof arr[0];
 
    // Function Call
    bool res = checkUniqueFrequency(arr, n);
 
    // Print the result
    if (res)
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
    return 0;
}

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Java

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// Java code for the above approach
import java.util.*;
 
class GFG{
 
// Function to check whether the
// frequency of elements in array
// is unique or not.
static boolean checkUniqueFrequency(int arr[],
                                    int n)
{
     
    // Freq map will store the frequency
    // of each element of the array
    HashMap<Integer,
            Integer> freq = new HashMap<Integer,
                                        Integer>();
 
    // Store the frequency of each
    // element from the array
    for(int i = 0; i < n; i++)
    {
        if(freq.containsKey(arr[i]))
        {
            freq.put(arr[i], freq.get(arr[i]) + 1);
        }else
        {
            freq.put(arr[i], 1);
        }
    }
 
    HashSet<Integer> uniqueFreq = new HashSet<Integer>();
 
    // Check whether frequency of any
    // two or more elements are same
    // or not. If yes, return false
    for(Map.Entry<Integer,
                  Integer> i : freq.entrySet())
    {
        if (uniqueFreq.contains(i.getValue()))
            return false;
        else
            uniqueFreq.add(i.getValue());
    }
 
    // Return true if each
    // frequency is unique
    return true;
}
 
// Driver Code
public static void main(String[] args)
{
     
    // Given array arr[]
    int arr[] = { 1, 1, 2, 5, 5 };
    int n = arr.length;
 
    // Function call
    boolean res = checkUniqueFrequency(arr, n);
 
    // Print the result
    if (res)
        System.out.print("Yes" + "\n");
    else
        System.out.print("No" + "\n");
}
}
 
// This code is contributed by PrinciRaj1992

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C#

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// C# code for the above approach
using System;
using System.Collections.Generic;
class GFG{
 
// Function to check whether the
// frequency of elements in array
// is unique or not.
static bool checkUniqueFrequency(int []arr,
                                 int n)
{
     
    // Freq map will store the frequency
    // of each element of the array
    Dictionary<int,
               int> freq = new Dictionary<int,
                                          int>();
 
    // Store the frequency of each
    // element from the array
    for(int i = 0; i < n; i++)
    {
        if(freq.ContainsKey(arr[i]))
        {
            freq[arr[i]] = freq[arr[i]] + 1;
        }else
        {
            freq.Add(arr[i], 1);
        }
    }
 
    HashSet<int> uniqueFreq = new HashSet<int>();
 
    // Check whether frequency of any
    // two or more elements are same
    // or not. If yes, return false
    foreach(KeyValuePair<int,
                         int> i in freq)
    {
        if (uniqueFreq.Contains(i.Value))
            return false;
        else
            uniqueFreq.Add(i.Value);
    }
 
    // Return true if each
    // frequency is unique
    return true;
}
 
// Driver Code
public static void Main(String[] args)
{
     
    // Given array []arr
    int []arr = { 1, 1, 2, 5, 5 };
    int n = arr.Length;
 
    // Function call
    bool res = checkUniqueFrequency(arr, n);
 
    // Print the result
    if (res)
        Console.Write("Yes" + "\n");
    else
        Console.Write("No" + "\n");
}
}
 
// This code is contributed by sapnasingh4991

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Output: 

No


Time Complexity: O(N), where N is the number of elements in the array.
Auxiliary Space: O(N) 

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