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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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

# 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 

Output:

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:

Time Complexity: O(n)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C++

Java

Python3

C#

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