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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)

Another Efficient Solution:

For large numbers when we should also work with modulus of 10^9+7. The above approach may not work. So the Intuition for this approach is for each number if we multiply it with prefix sum then all pair before the element are covered and then element is added to the sum which will guarantee us with the multiplication of pairs after the element.

Examples:

Input : A[] = {1, 3, 4, 5}
Output : 59
Possible Pairs : (1,3), (1,4), (1,5), (3,4), (3,5), (4,5)
Sum of Product : 1*3 + 1*4 + 1*5 + 3*4 + 3*5 +4*5 = 59
Intuition:
Initially ans=0, sum=0, i=0.so,
i=k :
    ans+=sum*A[k];
    sum+=A[k];
--------------------------------------
i=0 :A[i]=1
    ans+=(0)*(1);    ans==0
    sum+=1;            sum==1 (1)
i=1 : A[i]=3
    ans+=(1)*(3);    ans==3
    sum+=3;            sum==4 (1+3)
i=2 : A[i]=4
    ans+=(4)*(4);    ans==19
    sum+=4;            sum==8 (1+3+4)
i=3 : A[i]=5
    ans+=(8)*(5);    ans==59
    sum+=5;            sum==13 (1+3+4+5)
    
So, ans=59.

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)
{
    long long ans = 0;
    long long sum = 0;
    long long Mod = 1000000007;
    for (int i = 0; i < N; i++) {
        ans += (sum * A[i]) % Mod;
        ans %= Mod;
        sum += A[i];
        sum %= Mod;
    }
    return ans;
}
  
// 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;
}
  
// This code is contributed by Kasina Dheeraj.


Java




// Efficient Java program to find sum pair products in an
// array.
import java.io.*;
  
class GFG {
  
  // required function
  static int findProductSum(int[] A, int N)
  {
    long ans = 0;
    long sum = 0;
    long Mod = 1000000007;
    for (int i = 0; i < N; i++) {
      ans += (sum * A[i]) % Mod;
      ans %= Mod;
      sum += A[i];
      sum %= Mod;
    }
    return (int)ans;
  }
  
  public static void main(String[] args)
  {
    int[] A = { 1, 3, 4 };
    int n = A.length;
  
    System.out.print(
      "Sum of product of all pairs of array elements : "
      + findProductSum(A, n));
  }
}
  
// This code is contributed by lokeshmvs21.


Python3




# Efficient python program to find sum pair products in an array.
  
# Required function
def findProductSum(A, N):
    ans = 0
    Sum = 0
    Mod = 1000000007
    for i in range(N):
        ans += (Sum * A[i]) % Mod
        ans %= Mod
        Sum += A[i]
        Sum %= Mod
  
    return ans
  
A = [1, 3, 4]
n = len(A)
print("Sum of product of all pairs of array elements :", findProductSum(A, n))
  
# This code is contributed by lokeshmvs21.


C#




// Efficient C# program to find sum pair products in an
// array.
using System;
public class GFG 
{
  
  // required function
  static int findProductSum(int[] A, int N)
  {
    long ans = 0;
    long sum = 0;
    long Mod = 1000000007;
    for (int i = 0; i < N; i++) {
      ans += (sum * A[i]) % Mod;
      ans %= Mod;
      sum += A[i];
      sum %= Mod;
    }
    return (int)ans;
  }
  
  static public void Main()
  {
  
    // Code
    int[] A = { 1, 3, 4 };
    int n = A.Length;
  
    Console.Write(
      "Sum of product of all pairs of array elements : "
      + findProductSum(A, n));
  }
}
  
// This code is contributed by lokeshmvs21.


Javascript




class GFG
{
    // required function
    static findProductSum(A, N)
    {
        var ans = 0;
        var sum = 0;
        var Mod = 1000000007;
        for (var i=0; i < N; i++)
        {
            ans += (sum * A[i]) % Mod;
            ans %= Mod;
            sum += A[i];
            sum %= Mod;
        }
        return parseInt(ans);
    }
    static main(args)
    {
        var A = [1, 3, 4];
        var n = A.length;
        console.log("Sum of product of all pairs of array elements : " + GFG.findProductSum(A, n));
    }
}
GFG.main([]);
  
// This code is contributed by dhanshriborse561


Output:

Sum of product of all pairs of array elements : 19

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



Last Updated : 19 Sep, 2023
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