Sum of product of each element with each element after it

Last Updated : 11 Oct, 2022

Given an array arr[] of n integers. The task is to find the sum of product of each element with each element after it in the array. In other words, find sum of product of each arr[i] with each arr[j] such that j > i.

Examples :

Input : arr[] = {9, 3, 4, 2}
Output : 107
Explanation:
Sum of product of arr[0] with arr[1], arr[2], arr[3] is 9*3 + 9*4 + 9*2 = 81
Sum of product of arr[1] with arr[2], arr[3] is 3*4 + 3*2 = 18
Product of arr[2] with arr[3] is 4*2 = 8
Sum of all these sums is 81 + 18 + 8 = 107

Input : arr[] = {3, 5}
Output : 15

Observe to find (p*a + p*b + .....+ p*y + p*z) is equals to p*(a + b ..... y + z).
So for each i, 0 ? i ? n - 2 we have to calculate arr[i] * (arr[j] + arr[j+1] + ..... + arr[n - 2] + arr[n - 1]), where j = i + 1 and find sum of these.
We can reduce the complexity of finding (arr[j] + arr[j+1] + ..... + arr[n - 2] + arr[n - 1]) for each 'i' by precomputing the suffix sum of the array.
Lets sum[] stores the suffix sum of the given array i.e sum[i] have arr[i] + arr[i+1] + ..... + arr[n - 2] + arr[n - 1].
So, after precompute sum[], we need to find the sum of arr[i]*sum[i + 1] where 0 ? i ? n - 2.

Below is the implementation of this approach:

C++

// C++ Program to find sum of 
// product of each element
// with each element after it
#include <bits/stdc++.h>
using namespace std;

// Return sum of product 
// of each element with 
// each element after it
int findSumofProduct(int arr[], int n)
{
    int sum[n];
    sum[n - 1] = arr[n - 1];

    // finding suffix sum
    for (int i = n - 2; i >= 0; i--) 
        sum[i] = sum[i + 1] + arr[i];
    
    // finding the sum of product of 
    // each element with suffix sum.
    int res = 0;
    for (int i = 0; i < n - 1; i++) 
        res += (sum[i + 1] * arr[i]); 

    return res;
}

// Driver Code
int main()
{
    int arr[] = {9, 3, 4, 2};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSumofProduct(arr, n)
         << endl;
    return 0;
}

Java

// Java Program to find sum of product
// of each element with each element
// after it
class GFG {
    
    // Return sum of product of each
    // element with each element
    // after it
    static int findSumofProduct(int arr[],
                                    int n)
    {
        int sum[] = new int[n];
        sum[n - 1] = arr[n - 1];
    
        // finding suffix sum
        for (int i = n - 2; i >= 0; i--) 
            sum[i] = sum[i + 1] + arr[i];
        
        // finding the sum of product of
        // each element with suffix sum.
        int res = 0;
        for (int i = 0; i < n - 1; i++) 
            res += (sum[i + 1] * arr[i]); 
    
        return res;
    }
    
    // Driver Program
    public static void main(String[] args)
    {
        
        int arr[] = { 9, 3, 4, 2 };
        int n = arr.length;
        
        System.out.println(
                findSumofProduct(arr, n));
    }
}

// This code is contributed by
// Smitha Dinesh Semwal

Python3

# Python3 Program to find sum of 
# product of each element with 
# each element after it 

# Return sum of product of each 
# element with each element after it 
def findSumofProduct(arr, n):
    sum = [None for _ in range(n)]
    sum[n-1] = arr[n - 1]

    # finding suffix sum 
    for i in range(n - 2, -1, -1):
        sum[i] = sum[i + 1] + arr[i]

    # finding the sum of product of 
    # each element with suffix sum. 
    res = 0
    for i in range(n - 1):
        res += sum[i + 1] * arr[i]
    return res

# Driver Code
arr = [9, 3, 4, 2]
n = len(arr)
print(findSumofProduct(arr, n))

# This code is contributed
# by SamyuktaSHegde

// C# Program to find sum of product
// of each element with each element
// after it
using System;

class GFG {
    
    // Return sum of product of each
    // element with each element
    // after it
    static int findSumofProduct(int[] arr,
                                    int n)
    {
        int[] sum = new int[n];
        sum[n - 1] = arr[n - 1];
    
        // finding suffix sum
        for (int i = n - 2; i >= 0; i--) 
            sum[i] = sum[i + 1] + arr[i];
        
        // finding the sum of product of
        // each element with suffix sum.
        int res = 0;
        for (int i = 0; i < n - 1; i++) 
            res += (sum[i + 1] * arr[i]); 
    
        return res;
    }
    
    // Driver code
    static public void Main ()
    {
        int[] arr = { 9, 3, 4, 2 };
        int n = arr.Length;
        
        Console.WriteLine(findSumofProduct(arr, n));
    }
}

// This code is contributed by Ajit.

PHP

<?php
// PHP Program to find sum of 
// product of each element
// with each element after it

// Return sum of product 
// of each element with 
// each element after it
function findSumofProduct($arr, $n)
{
    
    $sum = array();
    $sum[$n - 1] = $arr[$n - 1];

    // finding suffix sum
    for ( $i = $n - 2; $i >= 0; $i--) 
        $sum[$i] = $sum[$i + 1] + $arr[$i];
    
    // finding the sum of product of 
    // each element with suffix sum.
    $res = 0;
    for ( $i = 0; $i < $n - 1; $i++) 
        $res += ($sum[$i + 1] * $arr[$i]); 

    return $res;
}

    // Driver Code
    $arr = array(9, 3, 4, 2);
    $n = count($arr);
    echo findSumofProduct($arr, $n);
    
// This code is contributed by anuJ_67.
?>

JavaScript

<script>

// Javascript Program to find sum of
// product of each element
// with each element after it

// Return sum of product
// of each element with
// each element after it
function findSumofProduct(arr, n)
{
    let sum = new Array(n);
    sum[n - 1] = arr[n - 1];

    // finding suffix sum
    for (let i = n - 2; i >= 0; i--)
        sum[i] = sum[i + 1] + arr[i];
    
    // finding the sum of product of
    // each element with suffix sum.
    let res = 0;
    for (let i = 0; i < n - 1; i++)
        res += (sum[i + 1] * arr[i]);

    return res;
}

// Driver Code

    let arr = [9, 3, 4, 2];
    let n = arr.length;
    document.write(findSumofProduct(arr, n)
        + "<br>");

// This code is contributed by Mayank Tyagi

</script>

Output

Time complexity: O(n) where n is the size of the given array
Auxiliary space: O(n)

Below is Optimized Code of finding sum of product of each element with each element after it:

C++

// C++ Program to find sum of
// product of each element 
// with each element after it
#include <bits/stdc++.h>
using namespace std;

// Return sum of product 
// of each element with 
// each element after it
int findSumofProduct(int arr[], 
                     int n)
{
    int suffix_sum = arr[n - 1];

    // Finding product of array 
    // elements and suffix sum.
    int res = 0;
    for (int i = n - 2; i >= 0; i--) 
    {

        res += (suffix_sum * arr[i]);

        // finding suffix sum
        suffix_sum += arr[i];
    }

    return res;
}

// Driver Code
int main()
{
    int arr[] = {9, 3, 4, 2};
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << findSumofProduct(arr, n) 
         << endl;
    return 0;
}

Java

// Java Program to find sum 
// of product of each 
// element with each
// element after it
import java .io.*;
class GFG {
    
    // Return sum of product of
    // each element with each
    // element after it
    static int findSumofProduct(int[] arr,
                                int n)
    {
        int suffix_sum = arr[n - 1];
    
        // Finding product of array 
        // elements and suffix sum.
        int res = 0;
        for (int i = n - 2; i >= 0; i--)
        {
            res += (suffix_sum * arr[i]);
    
            // finding suffix sum
            suffix_sum += arr[i];
        }
    
        return res;
    }

    // Driver code
    static public void main(String[] args)
    {
        int[] arr = {9, 3, 4, 2};
        int n = arr.length;
        System.out.println(findSumofProduct(arr, n));
    }
}

// This code is contributed by anuj_67.

Python 3

# Python 3 Program to find sum of
# product of each element 
# with each element after it

# Return sum of product 
# of each element with 
# each element after it
def findSumofProduct(arr, n):
    suffix_sum = arr[n - 1]

    # Finding product of array 
    # elements and suffix sum.
    res = 0
    for i in range(n - 2, -1, -1):

        res += (suffix_sum * arr[i])

        # finding suffix sum
        suffix_sum += arr[i]

    return res;

# Driver Code
arr = [9, 3, 4, 2];
n = len(arr);
print(findSumofProduct(arr, n))

# This code is contributed 
# by Akanksha Rai

// C# Program to find sum 
// of product of each 
// element with each
// element after it
using System;

class GFG {
    
    // Return sum of product of
    // each element with each
    // element after it
    static int findSumofProduct(int[] arr,
                                int n)
    {
        int suffix_sum = arr[n - 1];
    
        // Finding product of array 
        // elements and suffix sum.
        int res = 0;
        for (int i = n - 2; i >= 0; i--)
        {
            res += (suffix_sum * arr[i]);
    
            // finding suffix sum
            suffix_sum += arr[i];
        }
    
        return res;
    }

    // Driver code
    static public void Main()
    {
        int[] arr = {9, 3, 4, 2};
        int n = arr.Length;
        Console.WriteLine(findSumofProduct(arr, n));
    }
}

// This code is contributed by Ajit.

PHP

<?php
// PHP Program to find sum of
// product of each element 
// with each element after it

// Return sum of product 
// of each element with 
// each element after it
function findSumofProduct($arr, 
                          $n)
{
    $suffix_sum = $arr[$n - 1];

    // Finding product of array 
    // elements and suffix sum.
    $res = 0;
    for ( $i = $n - 2; $i >= 0; $i--) 
    {

        $res += ($suffix_sum * $arr[$i]);

        // finding suffix sum
        $suffix_sum += $arr[$i];
    }

    return $res;
}

    // Driver Code
    $arr = array(9, 3, 4, 2);
    $n = count($arr);
    echo findSumofProduct($arr, $n) 


// This code is contributed by anuJ_67.
?>

JavaScript

<script>

    // Javascript Program to find sum
    // of product of each
    // element with each
    // element after it
    
    // Return sum of product of
    // each element with each
    // element after it
    function findSumofProduct(arr, n)
    {
        let suffix_sum = arr[n - 1];
     
        // Finding product of array
        // elements and suffix sum.
        let res = 0;
        for (let i = n - 2; i >= 0; i--)
        {
            res += (suffix_sum * arr[i]);
     
            // finding suffix sum
            suffix_sum += arr[i];
        }
     
        return res;
    }
    
    let arr = [9, 3, 4, 2];
    let n = arr.length;
    document.write(findSumofProduct(arr, n));
    
</script>

Output

Time complexity: O(n) where n is the size of the given array
Auxiliary space: O(1)

Minimise sum of products of positions of each element with element at (i+K)%N

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Sum of product of each element with each element after it

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