# 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 ….., an-1, an}
Now, sum of product of all  possible pairs is =
{
(a1 * a1) + (a1 * a2) + (a1 * a3) + …..+ (a1 * an-1) + (a1, an) +
(a2 * a1) + (a2 * a2) + (a2 * a3) + ….. + (a2 * an-1) + (a2 * an) +
(a3 * a1) + (a3 * a2) + (a3 * a3) + ….. + (a3 * an-1) + (a3, an) +
…………………………………………………………………………..
(an, * a1) + (an * a2), +(an * a3) + ….. + (an * an-1) + (an, an)
}

={
(a1 + a2+  a3 + …..+ an-1 +  an * (a1 + a2+  a3 + …..+ an-1 +  an
}
=(a1 + a2+  a3 + …..+ an-1 +  an )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 ` `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); ` `    ``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`

Output:

```225

```

Time Complexity: O(N)

Auxiliary Space: O(1)

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