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Number of unique pairs in an array
  • Difficulty Level : Basic
  • Last Updated : 01 Feb, 2021

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:
(1, 1), (1, 2), (2, 1), (2, 2) are the only possible pairs.

Input: arr[] = {1, 2, 3} 
Output:

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++




// 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;
}

Java




// Java implementation of the approach
import java.awt.Point;
import java.util.*;
 
class GFG
{
 
// Function to return the number
// of unique pairs in the array
static int countUnique(int arr[], int n)
{
 
    // Set to store unique pairs
    Set<Point> s = new HashSet<>();
 
    // Make all possible pairs
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            s.add(new Point(arr[i], arr[j]));
 
    // Return the size of the set
    return s.size();
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 2, 4, 2, 5, 3, 5 };
    int n = arr.length;
 
    System.out.print(countUnique(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 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

C#




// C# implementation of the approach
using System;
using System.Collections;
using System.Collections.Generic;
 
class GFG{
     
public class store : IComparer<KeyValuePair<int, int>>
{
    public int Compare(KeyValuePair<int, int> x,
                       KeyValuePair<int, int> y)
    {
        if (x.Key != y.Key)
        {
            return x.Key.CompareTo(y.Key);   
        }
        else
        {
            return x.Value.CompareTo(y.Value);   
        }
    }
}
     
// Function to return the number
// of unique pairs in the array
static int countUnique(int []arr, int n)
{
     
    // Set to store unique pairs
    SortedSet<KeyValuePair<int,
                           int>> s = new  SortedSet<KeyValuePair<int,
                                                                 int>>(new store());
   
    // Make all possible pairs
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            s.Add(new KeyValuePair<int, int>(arr[i], arr[j]));
   
    // Return the size of the set
    return s.Count;
}
 
// Driver code   
public static void Main(string []arg)
{
    int []arr = { 1, 2, 2, 4, 2, 5, 3, 5 };
    int n = arr.Length;
     
    Console.Write(countUnique(arr, n));
}
}
 
// This code is contributed by rutvik_56
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++




// 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;
}

Java




// 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*/

Python3




# 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

C#




// 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
Output: 
25

 

Time Complexity: O(n) 

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