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:

Initialize the variable, res=0 to store the sum of array elements
Traverse the array, arr[] and add each element of the array to res.
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

Output:

Time Complexity: O(N)

Auxiliary Space: O(1)

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My Personal Notes

C++

Java

Python3

C#

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