Sum of all ordered pair-products from a given array

Given an array arr[] of size N, the task is to find the sum of all products of ordered pairs that can be generated from the given array elements.
Examples:

Input: arr[] ={1, 2, 3}
Output: 36
Explanation:All possible pairs are {(1, 1), {1, 2}, {1, 3}, {2, 1}, {2, 2}, {2, 3}, {3, 1}, {3, 2}, {3, 3}}. Therefore, the sum of product of all pairs is 36.

Input : arr[]={3, 4, 1, 2, 5}
Output: 225 

Naive Approach: The simplest approach is to iterate through all possible pairs from the given array calculate the sum of product of all pair-products. 

Time Complexity: O(N²)
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the following observations:

arr[]={a₁, a₂, a₃, a₄ ….., a_n-1, a_n}
Now, sum of product of all  possible pairs is =
{
(a_{1 *} a₁) + (a_{1 *} a₂) + (a_{1 *} a₃) + …..+ (a_{1 *} a_n-1) + (a₁, a_n) +
(a_{2 *} a₁) + (a_{2 *} a₂) + (a_{2 *} a₃) + ….. + (a_{2 *} a_n-1) + (a_{2 *} a_n) +
(a_{3 *} a₁) + (a_{3 *} a₂) + (a_{3 *} a₃) + ….. + (a_{3 *} a_n-1) + (a₃, a_n) +
…………………………………………………………………………..
(a_{n, *} a₁) + (a_{n *} a₂), +(a_{n *} a₃) + ….. + (a_{n *} a_n-1) + (a_n, a_n)
}

={
(a₁ + a₂+  a₃ + …..+ a_n-1 +  a_n ) _* (a₁ + a₂+  a₃ + …..+ a_n-1 +  a_n ) 
}
=(a₁ + a₂+  a₃ + …..+ a_n-1 +  a_n )²

Follow the steps below to solve the problem: 

1. Initialize the variable, res=0 to store the sum of array elements
2. Traverse the array, arr[] and add each element of the array to res.
3. Finally, print the square of the res as the required answer.

Below is the implementation of the above approach:

C++

// C++ program to implement
// the above approach

#include <bits/stdc++.h>
using namespace std;

// Function to calculate the
// sum of all pair-products
int sumOfProd(int arr[], int N)
{
    // Stores sum of array
    int sum = 0;

    for (int i = 0; i < N; i++) {

        // Update sum of the array
        sum += arr[i];
    }

    return sum * sum;
}

// Driver Code
int main()
{
    int arr[] = { 2, 3, 1, 5, 4 };
    int N = sizeof(arr) / sizeof(arr[0]);
    cout << sumOfProd(arr, N);
    return 0;
}

Java

// Java program to implement
// the above approach
import java.util.*;

class GFG{

// Function to calculate the
// sum of all pair-products
static int sumOfProd(int arr[], int N)
{

    // Stores sum of array
    int sum = 0;

    for(int i = 0; i < N; i++)
    {

        // Update sum of the array
        sum += arr[i];
    }
    return sum * sum;
}

// Driver code
public static void main(String[] args)
{
    int arr[] = { 2, 3, 1, 5, 4 };
    int N = arr.length;

    System.out.print(sumOfProd(arr, N));
}
}

// This code is contributed by offbeat

Python3

# Python3 program to implement
# the above approach

# Function to calculate the
# sum of all pair-products
def sumOfProd(arr, N):

    # Stores sum of array
    sum = 0

    for i in range(N):

        # Update sum of the array
        sum += arr[i]

    return sum * sum

# Driver Code
if __name__ == '__main__':

    arr = [ 2, 3, 1, 5, 4 ]
    N = len(arr)

    print(sumOfProd(arr, N))

# This code is contributed by SURENDRA_GANGWAR

C#

// C# program to implement
// the above approach
using System;

class GFG{

// Function to calculate the
// sum of all pair-products
static int sumOfProd(int[] arr, int N)
{

    // Stores sum of array
    int sum = 0;

    for(int i = 0; i < N; i++)
    {

        // Update sum of the array
        sum += arr[i];
    }
    return sum * sum;
}

// Driver code
static void Main()
{
    int[] arr = { 2, 3, 1, 5, 4 };
    int N = arr.Length;

    Console.WriteLine(sumOfProd(arr, N));
}
}

// This code is contributed by divyeshrabadiya07

Javascript

<script>

// JavaScript program for the above approach

// Function to calculate the
// sum of all pair-products
function sumOfProd(arr, N)
{

    // Stores sum of array
    let sum = 0;

    for(let i = 0; i < N; i++)
    {

        // Update sum of the array
        sum += arr[i];
    }
    return sum * sum;
}

// Driver Code

       let arr = [ 2, 3, 1, 5, 4 ];
    let N = arr.length;

    document.write(sumOfProd(arr, N));

// This code is contributed by souravghosh0416.
</script>

Output:

225

Time Complexity: O(N)

Auxiliary Space: O(1)