Count of distinct pair sum in given Array

Given an array arr[] of size N, the task is to find the total number of unique pair sums possible from the array elements.

Examples:

Input: arr[] = {6, 1, 4, 3}
Output: 5
Explanation: All pair possible are {6, 1}, {6, 4}, {6, 3}, {1, 4}, {1, 3}, {4, 3}. S
ums of these pairs are 7, 10, 9, 5, 4, 7. So unique sums 7, 10, 9, 5, 4. So answer is 5.

Input: arr[] = {8, 7, 6, 5, 4, 3, 2, 1}
Output: 13

Approach: This problem can be efficiently solved by using unordered_set. 

Calculate all possible sum of pairs and store them in an unordered set. This is done to store the store elements in an average time of O(1) with no value repeating. 

Algorithm:

• Use nested loops to get all possible sums of elements of the array.
• Store all the possible sums into an unordered_set.
• Total possible unique sums would be equal to the size of unordered_set. So return the size of the unordered set.

Below is the implementation of the above approach:

C++

// C++ program to count the pairs with unique sums
#include <bits/stdc++.h>
using namespace std;

// Function to return the
// total count of required pairs
int count(int arr[], int n)
{
    unordered_set<int> s;
    // Add all possible sums into the set
    for (int i = 0; i < n - 1; i++) {
        for (int j = i + 1; j < n; j++) {
            s.insert(arr[i] + arr[j]);
        }
    }
    // Return the size of set
    return s.size();
}

// Driver code
int main()
{
    int arr[] = { 6, 1, 4, 3 };
    int N = sizeof(arr) / sizeof(arr[0]);

    // Function call
    cout << count(arr, N);
    return 0;
}

Java

// Java program to find Count the pairs
// with unique sums
import java.util.*;

class GFG {
    // Function to return the
    // total count of required pairs
    static int count(int arr[], int n)
    {
        Set<Integer> s = new HashSet<Integer>();

        // Add all possible sums into the set
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                s.add(arr[i] + arr[j]);
            }
        }
        // Return the size of set
        return s.size();
    }

    // Driver code
    public static void main(String args[])
    {
        int arr[] = { 6, 1, 4, 3 };
        int N = arr.length;

        // Function call
        System.out.println(count(arr, N));
    }
}

Python3

# Python program to find Count the pairs
# with unique sums

# Function to return the
# total count of required pairs
def count(arr, n):
    s = set()

    # Add all possible sums into the set
    for i in range(n-1):
        for j in range(i + 1, n):
            s.add(arr[i] + arr[j])

    # Return the size of set
    return int(len(s))

# Driver code
if __name__ == '__main__':
    arr = [6, 1, 4, 3]
    N = len(arr)

    # Function call
    print(count(arr, N))

C#

// C# program to count the pairs with unique sums
using System;
using System.Collections.Generic;

public class GFG {

    // Function to return the
    // total count of required pairs
    static int count(int []arr, int n)
    {
        HashSet<int> s = new HashSet<int>();

        // Add all possible sums into the set
        for (int i = 0; i < n - 1; i++) {
            for (int j = i + 1; j < n; j++) {
                s.Add(arr[i] + arr[j]);
            }
        }
        // Return the size of set
        return s.Count;
    }

    // Driver code
    static public void Main()
    {
        int []arr = { 6, 1, 4, 3 };
        int N = arr.Length;

        // Function call
        Console.WriteLine(count(arr, N));
    }
}

// This code is contributed by Rohit Pradhan

Javascript

<script>

// JavaScript program to find Count the pairs
// with unique sums

// Function to return the
// total count of required pairs
function count(arr, n) {
    let s = new Set();
    // Add all possible sums into the set
    for (let i = 0; i < n - 1; i++) {
        for (let j = i + 1; j < n; j++) {
            s.add(arr[i] + arr[j]);
        }
    }
    // Return the size of set
    return s.size;
}

// Driver code
let arr = [6, 1, 4, 3];
let N = arr.length;

// Function call
document.write(count(arr, N));

</script>

Output

5

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