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Find the sum of all possible pairs in an array of N elements
  • Difficulty Level : Easy
  • Last Updated : 15 Oct, 2019

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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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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++

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// 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;
}

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Java

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// 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

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Python3

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# 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

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C#

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// 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

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Output:

36

Time Complexity: O(N)

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