Number of unique pairs in an array

Given an array of N elements, the task is to find all the unique pairs that can be formed using the elements of a given array.

Examples:

Input: arr[] = {1, 1, 2}
Output: 4
(1, 1), (1, 2), (2, 1), (2, 2) are the only possible pairs.

Input: arr[] = {1, 2, 3}
Output: 9

Naive approach: The simple solution is to iterate through every possible pair and add them to a set and then find out the size of the set.

Below is the implementation of the above approach:

C++

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to return the number
// of unique pairs in the array
int countUnique(int arr[], int n)
{
  
    // Set to store unique pairs
    set<pair<int, int> > s;
  
    // Make all possible pairs
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            s.insert(make_pair(arr[i], arr[j]));
  
    // Return the size of the set
    return s.size();
}
  
// Driver code
int main()
{
    int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
    int n = sizeof(arr) / sizeof(arr[0]);
  
    cout << countUnique(arr, n);
    return 0;
}

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Python














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# Python implementation of the approach 


  


# Function to return the number 


# of unique pairs in the array 


def countUnique(arr, n): 


    # Set to store unique pairs 


    s = set() 


  


    # Make all possible pairs 


    for i in range(n): 


        for j in range(n): 


            s.add((arr[i], arr[j])) 


  


    # Return the size of the set 


    return len(s) 


  


  


# Driver code 


  


arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ] 


n = len(arr) 


print(countUnique(arr, n)) 


  


# This code is contributed by ankush_953 



















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Output:


25





Time Complexity: Time complexity of the above implementation is O(n2 Log n). We can optimize it to O(n2) using unordered_set with user defined hash function.


Efficient approach: First find out the number of unique elements in an array. Let the number of unique elements be x. Then, the number of unique pairs would be x2. This is because each unique element can form a pair with every other unique element including itself.


Below is the implementation of the above approach:




C++















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// C++ implementation of the approach 


#include <bits/stdc++.h> 


using namespace std; 


  


// Function to return the number 


// of unique pairs in the array 


int countUnique(int arr[], int n) 


{ 


  


    unordered_set<int> s; 


    for (int i = 0; i < n; i++) 


        s.insert(arr[i]); 


  


    int count = pow(s.size(), 2); 


  


    return count; 


} 


  


// Driver code 


int main() 


{ 


    int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 }; 


    int n = sizeof(arr) / sizeof(arr[0]); 


  


    cout << countUnique(arr, n); 


    return 0; 


} 



















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Java














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// Java implementation of the approach 


import java.util.*; 


  


class GFG  


{ 


  


    // Function to return the number 


    // of unique pairs in the array 


    static int countUnique(int arr[], int n)  


    { 


  


        HashSet<Integer> s = new HashSet<>(); 


        for (int i = 0; i < n; i++) 


        { 


            s.add(arr[i]); 


        } 


        int count = (int) Math.pow(s.size(), 2); 


  


        return count; 


    } 


  


    // Driver code 


    public static void main(String[] args) 


    { 


        int arr[] = {1, 2, 2, 4, 2, 5, 3, 5}; 


        int n = arr.length; 


        System.out.println(countUnique(arr, n)); 


    } 


} 


  


/* This code has been contributed  


by PrinciRaj1992*/



















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Python3














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# Python3 implementation of the approach  


  


# Function to return the number  


# of unique pairs in the array  


def countUnique(arr, n) : 


      


    s = set();  


    for i in range(n) : 


        s.add(arr[i]);  


  


    count = pow(len(s), 2);  


  


    return count;  


  


# Driver code  


if __name__ == "__main__" :  


  


    arr = [ 1, 2, 2, 4, 2, 5, 3, 5 ];  


    n = len(arr); 


  


    print(countUnique(arr, n)); 


      


# This code is contributed by Ryuga 



















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C#














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// C# implementation of the approach 


using System;  


using System.Collections.Generic;  


  


class GFG  


{ 


  


    // Function to return the number 


    // of unique pairs in the array 


    static int countUnique(int []arr, int n)  


    { 


  


        HashSet<int> s = new HashSet<int>(); 


        for (int i = 0; i < n; i++) 


        { 


            s.Add(arr[i]); 


        } 


        int count = (int) Math.Pow(s.Count, 2); 


  


        return count; 


    } 


  


    // Driver code 


    static void Main() 


    { 


        int []arr = {1, 2, 2, 4, 2, 5, 3, 5}; 


        int n = arr.Length; 


        Console.WriteLine(countUnique(arr, n)); 


    } 


} 


  


// This code has been contributed by mits 



















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Output:


25





Time Complexity: O(n)



          
          
          
          

            

    

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