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Find the sum of all possible pairs in an array of N elements

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  • Difficulty Level : Easy
  • Last Updated : 19 Mar, 2022
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Given an array arr[] of N integers, the task is to find the sum of all the pairs possible from the given array. Note that, 

  1. (arr[i], arr[i]) is also considered as a valid pair.
  2. (arr[i], arr[j]) and (arr[j], arr[i]) are considered as two different pairs.

Examples: 

Input: arr[] = {1, 2} 
Output: 12 
All valid pairs are (1, 1), (1, 2), (2, 1) and (2, 2). 
1 + 1 + 1 + 2 + 2 + 1 + 2 + 2 = 12

Input: arr[] = {1, 2, 3, 1, 4} 
Output: 110 
 

Naive approach: Find all the possible pairs and calculate the sum of the elements of each pair.

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 sum of the elements
// of all possible pairs from the array
int sumPairs(int arr[], int n)
{
 
    // To store the required sum
    int sum = 0;
 
    // Nested loop for all possible pairs
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
 
            // Add the sum of the elements
            // of the current pair
            sum += (arr[i] + arr[j]);
        }
    }
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << sumPairs(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
 
class GFG
{
 
    // Function to return the sum of the elements
    // of all possible pairs from the array
    static int sumPairs(int arr[], int n)
    {
 
        // To store the required sum
        int sum = 0;
 
        // Nested loop for all possible pairs
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
 
                // Add the sum of the elements
                // of the current pair
                sum += (arr[i] + arr[j]);
            }
        }
        return sum;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int arr[] = {1, 2, 3};
        int n = arr.length;
 
        System.out.println(sumPairs(arr, n));
    }
}
 
// This code is contributed by PrinciRaj1992

Python3




# Python3 implementation of the approach
 
# Function to return the summ of the elements
# of all possible pairs from the array
def summPairs(arr, n):
 
    # To store the required summ
    summ = 0
 
    # Nested loop for all possible pairs
    for i in range(n):
        for j in range(n):
 
            # Add the summ of the elements
            # of the current pair
            summ += (arr[i] + arr[j])
 
    return summ
 
# Driver code
arr = [1, 2, 3]
n = len(arr)
 
print(summPairs(arr, n))
 
# This code is contributed by Mohit Kumar

C#




// C# implementation of the approach
using System;
 
class GFG
{
 
    // Function to return the sum of the elements
    // of all possible pairs from the array
    static int sumPairs(int []arr, int n)
    {
 
        // To store the required sum
        int sum = 0;
 
        // Nested loop for all possible pairs
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
 
                // Add the sum of the elements
                // of the current pair
                sum += (arr[i] + arr[j]);
            }
        }
        return sum;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int []arr = {1, 2, 3};
        int n = arr.Length;
 
        Console.WriteLine(sumPairs(arr, n));
    }
}
     
// This code is contributed by PrinciRaj1992

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the sum of the elements
// of all possible pairs from the array
function sumPairs(arr, n)
{
     
    // To store the required sum
    var sum = 0;
    var i, j;
     
    // Nested loop for all possible pairs
    for(i = 0; i < n; i++)
    {
        for(j = 0; j < n; j++)
        {
             
            // Add the sum of the elements
            // of the current pair
            sum += (arr[i] + arr[j]);
        }
    }
    return sum;
}
 
// Driver code
var arr = [ 1, 2, 3 ];
var n = arr.length;
 
document.write(sumPairs(arr, n));
 
// This code is contributed by ipg2016107
 
</script>

Output: 

36

 

Time Complexity: O(N2)

Auxiliary Space: O(1)

Efficient approach: It can be observed that each element appears exactly (2 * N) times as one of the elements of the pair (x, y). Exactly N times as x and exactly N times as y.

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 sum of the elements
// of all possible pairs from the array
int sumPairs(int arr[], int n)
{
 
    // To store the required sum
    int sum = 0;
 
    // For every element of the array
    for (int i = 0; i < n; i++) {
 
        // It appears (2 * n) times
        sum = sum + (arr[i] * (2 * n));
    }
 
    return sum;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
 
    cout << sumPairs(arr, n);
 
    return 0;
}

Java




// Java implementation of the approach
import java.util.*;
     
class GFG
{
 
// Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs(int arr[], int n)
{
 
    // To store the required sum
    int sum = 0;
 
    // For every element of the array
    for (int i = 0; i < n; i++)
    {
 
        // It appears (2 * n) times
        sum = sum + (arr[i] * (2 * n));
    }
 
    return sum;
}
 
// Driver code
static public void main(String []arg)
{
    int arr[] = { 1, 2, 3 };
    int n = arr.length;
 
    System.out.println(sumPairs(arr, n));
}
}
 
// This code is contributed by 29AjayKumar

Python3




# Python3 implementation of the approach
 
# Function to return the sum of the elements
# of all possible pairs from the array
def sumPairs(arr, n) :
 
    # To store the required sum
    sum = 0;
 
    # For every element of the array
    for i in range(n) :
 
        # It appears (2 * n) times
        sum = sum + (arr[i] * (2 * n));
 
    return sum;
 
# Driver code
if __name__ == "__main__" :
 
    arr = [ 1, 2, 3 ];
    n = len(arr);
 
    print(sumPairs(arr, n));
 
# This code is contributed by AnkitRai01

C#




// C# implementation of the approach
using System;        
 
class GFG
{
 
// Function to return the sum of the elements
// of all possible pairs from the array
static int sumPairs(int []arr, int n)
{
 
    // To store the required sum
    int sum = 0;
 
    // For every element of the array
    for (int i = 0; i < n; i++)
    {
 
        // It appears (2 * n) times
        sum = sum + (arr[i] * (2 * n));
    }
 
    return sum;
}
 
// Driver code
static public void Main(String []arg)
{
    int []arr = { 1, 2, 3 };
    int n = arr.Length;
 
    Console.WriteLine(sumPairs(arr, n));
}
}
 
// This code contributed by Rajput-Ji

Javascript




<script>
 
// Javascript implementation of the approach
 
// Function to return the sum of the elements
// of all possible pairs from the array
function sumPairs(arr, n)
{
 
    // To store the required sum
    let sum = 0;
 
    // For every element of the array
    for (let i = 0; i < n; i++) {
 
        // It appears (2 * n) times
        sum = sum + (arr[i] * (2 * n));
    }
 
    return sum;
}
 
// Driver code
    let arr = [ 1, 2, 3 ];
    let n = arr.length;
 
    document.write(sumPairs(arr, n));
 
</script>

Output: 

36

 

Time Complexity: O(N)
 Auxiliary Space: O(1)


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