Skip to content
Related Articles

Related Articles

Improve Article

Sum of product of all pairs of array elements

  • Difficulty Level : Easy
  • Last Updated : 14 Aug, 2021

Given an array A[] of integers find sum of product of all pairs of array elements i. e., we need to find of product after execution of following pseudo code 

product = 0
for i = 1:n
    for j = i+1:n
        product = product + A[i]*A[j]

Examples: 

Input : A[] = {1, 3, 4}
Output : 19
Possible Pairs : (1,3), (1,4), (3,4)
Sum of Product : 1*3 + 1*4 + 3*4 = 19

 

 

Naive Solution :

For each index i we loop through j=i+1 to j=n and add A[i]*A[j] each time. Below is implementation for the same. 
 



C++




// A naive C++ program to find sum of product
#include <iostream>
using namespace std;
 
// Returns sum of pair products
int findProductSum(int A[], int n)
{
    int product = 0;
    for (int i = 0; i < n; i++)
        for (int j = i+1; j < n; j++)
            product = product + A[i]*A[j];
    return product;
}
 
// Driver code
int main()
{
    int A[] = {1, 3, 4};
    int n = sizeof(A)/sizeof(A[0]);
 
    cout << "sum of product of all pairs "
    "of array elements : " << findProductSum(A, n);
 
    return 0;
}

Java




/*package whatever //do not write package name here */
// A naive Java program to find sum of product
import java.io.*;
class test
{
    // Returns sum of pair products
    int findProductSum(int A[], int n)
    {
    int product = 0;
    for (int i = 0; i < n; i++)
        for (int j = i+1; j < n; j++)
            product = product + A[i]*A[j];
    return product;
    }
}
class GFG {
 
// Driver code
    public static void main (String[] args) {
 
    int A[] = {1, 3, 4};
    int n = A.length;
    test t = new test();
    System.out.print("sum of product of all pairs of array elements : ");
    System.out.println(t.findProductSum(A, n));
 
    }
}

Python3




# A naive python3 program to find sum of product
  
# Returns sum of pair products
def findProductSum(A,n):
 
    product = 0
    for i in range (n):
        for j in range ( i+1,n):
            product = product + A[i]*A[j]
    return product
  
# Driver code
if __name__=="__main__":
 
    A = [1, 3, 4]
    n = len (A)
  
    print("sum of product of all pairs "
    "of array elements : " ,findProductSum(A, n))

C#




// A naive C# program to find sum of product
using System;
 
class GFG
{
     
// Returns sum of pair products
static int findProductSum(int[] A, int n)
{
    int product = 0;
    for (int i = 0; i < n; i++)
        for (int j = i + 1; j < n; j++)
            product = product + A[i] * A[j];
    return product;
}
 
// Driver code
public static void Main()
{
    int[] A = {1, 3, 4};
    int n = A.Length;
    Console.WriteLine("sum of product of all " +
                      "pairs of array elements : ");
    Console.WriteLine(findProductSum(A, n));
}
}
 
// This code is contributed
// by Akanksha Rai

PHP




<?php
// A naive PHP program to find
// sum of product
 
// Returns sum of pair products
function findProductSum($A, $n)
{
    $product = 0;
    for ($i = 0; $i < $n; $i++)
        for ($j = $i + 1; $j < $n; $j++)
            $product = $product + $A[$i] * $A[$j];
    return $product;
}
 
    // Driver code
    $A = array (1, 3, 4);
    $n = sizeof($A);
 
    echo "sum of product of all pairs ",
         "of array elements : "
         ,findProductSum($A, $n);
 
// This code is contributed by aj_36
?>

Javascript




<script>
 
// A naive Javascript program to find sum of product
   
// Returns sum of pair products
function findProductSum(A, n)
{
    let product = 0;
    for (let i= 0; i < n; i++)
        for (let j = i+1; j < n; j++)
            product = product + A[i]*A[j];
    return product;
}
   
// Driver code
    let A = [1, 3, 4];
    let n = A.length;
   
    document.write("sum of product of all pairs " +
    "of array elements : " + findProductSum(A, n));
     
// This code is contributed by Mayank Tyagi
   
</script>

Output: 

sum of product of all pairs of array elements : 19

Time Complexity : O(n2
Space Complexity : O(1)
 

Efficient O(n) solution :

We know that
(a + b + c)2 = a2 + b2 + c2 + 2*(a*b + b*c + c*a)
Let required sum be P
Let E = (a1 + a2 + a3 + a4 ... + an)^2 
=> E = a12 + a22 + ... + an2 + 2*(a1*a2 + a1*a3 + ....)
=> E = a12 + a22 + ... + an2 + 2*(P)
=> P = ( E - (a12 + a22 + .... + an2) ) / 2

 

C++




// Efficient C++ program to find sum pair products
// in an array.
#include <iostream>
using namespace std;
 
// required function
int findProductSum(int A[], int n)
{
    // calculating array sum (a1 + a2  ... + an)
    int array_sum = 0;
    for (int i = 0; i < n; i++)
        array_sum = array_sum + A[i];
 
    // calculating square of array sum
    // (a1 + a2 + ... + an)^2
    int array_sum_square = array_sum * array_sum;
 
    // calculating a1^2 + a2^2 + ... + an^2
    int individual_square_sum = 0;
    for (int i = 0; i < n; i++)
        individual_square_sum += A[i]*A[i];
 
    // required sum is (array_sum_square -
    // individual_square_sum) / 2
    return (array_sum_square - individual_square_sum)/2;
}
 
// Driver code
int main()
{
    int A[] = {1, 3, 4};
    int n = sizeof(A)/sizeof(A[0]);
    cout << "sum of product of all pairs of array "
            "elements : " << findProductSum(A, n);
    return 0;
}

Java




// Efficient Java program to find sum pair products
// in an array.
class GFG
{
 
// required function
static int findProductSum(int A[], int n)
{
    // calculating array sum (a1 + a2 ... + an)
    int array_sum = 0;
    for (int i = 0; i < n; i++)
        array_sum = array_sum + A[i];
 
    // calculating square of array sum
    // (a1 + a2 + ... + an)^2
    int array_sum_square = array_sum * array_sum;
 
    // calculating a1^2 + a2^2 + ... + an^2
    int individual_square_sum = 0;
    for (int i = 0; i < n; i++)
        individual_square_sum += A[i] * A[i];
 
    // required sum is (array_sum_square -
    // individual_square_sum) / 2
    return (array_sum_square - individual_square_sum) / 2;
}
 
// Driver code
public static void main(String[] args)
{
    int A[] = {1, 3, 4};
    int n = A.length;
    System.out.println("sum of product of all pairs of array "
            +"elements : " + findProductSum(A, n));
    }
}
 
// This code is contributed by 29AjayKumar

Python3




# Efficient python 3 program to find sum
# pair products in an array.
 
# required function
def findProductSum(A, n):
     
    # calculating array sum (a1 + a2 ... + an)
    array_sum = 0
    for i in range(0, n, 1):
        array_sum = array_sum + A[i]
 
    # calculating square of array sum
    # (a1 + a2 + ... + an)^2
    array_sum_square = array_sum * array_sum
 
    # calculating a1^2 + a2^2 + ... + an^2
    individual_square_sum = 0
    for i in range(0, n, 1):
        individual_square_sum += A[i] * A[i]
 
    # required sum is (array_sum_square -
    # individual_square_sum) / 2
    return (array_sum_square -
            individual_square_sum) / 2
 
# Driver code
if __name__ == '__main__':
    A = [1, 3, 4]
    n = len(A)
    print("sum of product of all pairs of",
          "array elements :", int(findProductSum(A, n)))
     
# This code is contributed by
# Sahil_Shelangia

C#




// Efficient C# program to find sum pair
// products in an array.
using System;
 
class GFG
{
 
// required function
static int findProductSum(int[] A, int n)
{
    // calculating array sum (a1 + a2 ... + an)
    int array_sum = 0;
    for (int i = 0; i < n; i++)
        array_sum = array_sum + A[i];
 
    // calculating square of array sum
    // (a1 + a2 + ... + an)^2
    int array_sum_square = array_sum * array_sum;
 
    // calculating a1^2 + a2^2 + ... + an^2
    int individual_square_sum = 0;
    for (int i = 0; i < n; i++)
        individual_square_sum += A[i] * A[i];
 
    // required sum is (array_sum_square -
    // individual_square_sum) / 2
    return (array_sum_square -
            individual_square_sum) / 2;
}
 
// Driver code
public static void Main()
{
    int[] A = {1, 3, 4};
    int n = A.Length;
    Console.WriteLine("sum of product of all " +
                      "pairs of array elements : " +
                       findProductSum(A, n));
}
}
 
// This code is contributed by Akanksha Rai

PHP




<?php
// Efficient PHP program to find sum
// pair products in an array.
 
// required function
function findProductSum(&$A, $n)
{
    // calculating array sum (a1 + a2 ... + an)
    $array_sum = 0;
    for ($i = 0; $i < $n; $i++)
        $array_sum = $array_sum + $A[$i];
 
    // calculating square of array sum
    // (a1 + a2 + ... + an)^2
    $array_sum_square = $array_sum * $array_sum;
 
    // calculating a1^2 + a2^2 + ... + an^2
    $individual_square_sum = 0;
    for ($i = 0; $i < $n; $i++)
        $individual_square_sum += $A[$i] * $A[$i];
 
    // required sum is (array_sum_square -
    // individual_square_sum) / 2
    return ($array_sum_square -
            $individual_square_sum) / 2;
}
 
// Driver code
$A = array(1, 3, 4);
$n = sizeof($A);
echo("sum of product of all pairs " .
             "of array elements : ");
echo (findProductSum($A, $n));
 
// This code is contributed
// by Shivi_Aggarwal
?>

Javascript




<script>
    // Efficient Javascript program to find sum pair
    // products in an array.
     
    // required function
    function findProductSum(A, n)
    {
        // calculating array sum (a1 + a2 ... + an)
        let array_sum = 0;
        for (let i = 0; i < n; i++)
            array_sum = array_sum + A[i];
 
        // calculating square of array sum
        // (a1 + a2 + ... + an)^2
        let array_sum_square = array_sum * array_sum;
 
        // calculating a1^2 + a2^2 + ... + an^2
        let individual_square_sum = 0;
        for (let i = 0; i < n; i++)
            individual_square_sum += A[i] * A[i];
 
        // required sum is (array_sum_square -
        // individual_square_sum) / 2
        return (array_sum_square - individual_square_sum) / 2;
    }
     
    let A = [1, 3, 4];
    let n = A.length;
    document.write("sum of product of all " +
                      "pairs of array elements : " +
                       findProductSum(A, n));
     
    // This code is contributed by rameshtravel07.
</script>

Output: 

sum of product of all pairs of array elements : 19

Time Complexity : O(n) 
Space Complexity : O(1)
This article is contributed by Pratik Chhajer. If you like GeeksforGeeks and would like to contribute, you can also write an article using write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.




My Personal Notes arrow_drop_up
Recommended Articles
Page :