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

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

36

Time Complexity: O(N2)

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# 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

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


Output:

36

Time Complexity: O(N)



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.