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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 ``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 > 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 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>``{``    ``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> s = ``new`  `SortedSet>(``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 ``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 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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