Find maximum value of Sum( i*arr[i]) with only rotations on given array allowed

Given an array arr[] of size N, the task is to find the maximum possible sum of i*arr[i] when the array can be rotated any number of times.

Examples :

Input: arr[] = {1, 20, 2, 10}
Output: 72.We can get 72 by rotating array twice.
{2, 10, 1, 20}
20*3 + 1*2 + 10*1 + 2*0 = 72

Input: arr[] = {10, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Output: 330
We can get 330 by rotating array 9 times.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
0*1 + 1*2 + 2*3 … 9*10 = 330

Naive Approach: The basic idea of this approach is 

Find all rotations one by one, check the sum of every rotation and return the maximum sum. 

Algorithm

• Start by initializing max_sum to INT_MIN
• Loop say i,from 0 to n-1:
  • a. Initialize sum to 0
  • b. Loop j from 0 to n-1:
    • i. Calculate the index of the j-th element after rotation: (i+j) % n
    • ii. Add the product of the element and its index to sum: j * arr[(i+j) % n]
  • c. If sum is greater than max_sum, update max_sum to sum
• Return max_sum

C++

#include <climits>
#include <iostream>

using namespace std;

int max_sum_rotation(int arr[], int n)
{
    int max_sum = INT_MIN; // set the maximum sum to the
                           // minimum possible value
    for (int i = 0; i < n;
         i++) { // loop through all possible rotations
        int sum = 0; // set the current sum to zero
        for (int j = 0; j < n;
             j++) { // loop through all elements in the
                    // array
            int index
                = (i + j)
                  % n; // calculate the index of the current
                       // element after rotation
            sum += j * arr[index]; // add the product of the
                                   // element with its index
                                   // to the sum
        }
        max_sum = max(
            max_sum,
            sum); // update the maximum sum if necessary
    }
    return max_sum; // return the maximum sum obtained over
                    // all rotations
}

int main()
{
    int arr[] = {
        10, 1, 2, 3, 4, 5, 6, 7, 8, 9
    }; // define an array
    int n = sizeof(arr)
            / sizeof(
                arr[0]); // calculate the size of the array
    cout << max_sum_rotation(arr, n)
         << endl; // call the function and print the result
    return 0; // indicate successful program completion
}

Output

330

Time Complexity: O(N²), where n is the size of the input array. 
Auxiliary Space: O(1), because it uses a constant amount of extra space to store the variables max_sum, sum, i, and j. 

Efficient Approach: The idea is as follows:

Let R_j be value of i*arr[i] with j rotations. 

• The idea is to calculate the next rotation value from the previous rotation, i.e., calculate R_j from R_j-1. 
• We can calculate the initial value of the result as R₀, then keep calculating the next rotation values. 

How to efficiently calculate R_j from R_j-1?

This can be done in O(1) time. Below are the details. 

Let us calculate initial value of i*arr[i] with no rotation
R₀ = 0*arr[0] + 1*arr[1] +…+ (n-1)*arr[n-1]

After 1 rotation arr[n-1], becomes first element of array, 

• arr[0] becomes second element, arr[1] becomes third element and so on.
• R₁ = 0*arr[n-1] + 1*arr[0] +…+ (n-1)*arr[n-2]
• R₁ – R₀ = arr[0] + arr[1] + … + arr[n-2] – (n-1)*arr[n-1]

After 2 rotations arr[n-2], becomes first element of array, 

• arr[n-1] becomes second element, arr[0] becomes third element and so on.
• R₂ = 0*arr[n-2] + 1*arr[n-1] +…+ (n-1)*arr[n-3]
• R₂ – R₁ = arr[0] + arr[1] + … + arr[n-3] – (n-1)*arr[n-2] + arr[n-1]

If we take a closer look at above values, we can observe below pattern
R_j – R_j-1 = arrSum – n * arr[n-j],
Where arrSum is sum of all array elements, i.e.,  arrSum = ∑ arr[i] , 0 ≤ i ≤ N-1

Follow the below illustration for a better understanding. 

Illustration:

Given arr[]={10, 1, 2, 3, 4, 5, 6, 7, 8, 9}, 

arrSum = 55, currVal = summation of (i*arr[i]) =  285
In each iteration the currVal is currVal = currVal + arrSum-n*arr[n-j] ,

1st rotation: currVal = 285 + 55 –  (10 *  9) =  250

2nd rotation: currVal = 285 + 55 – (10 * 8) = 260

3rd rotation: currVal = 285 + 55 – (10 * 7) = 270
.
.
.
Last rotation: currVal = 285 + 55 – (10 * 1) = 330

Previous currVal was 285, now it becomes 330.
It’s the maximum value we can find hence return 330.

Follow the steps mentioned below to implement the above approach:

• Compute the sum of all array elements. Let this sum be ‘arrSum‘.
• Compute R₀ for the given array. Let this value be currVal.
• Loop from j = 1 to N-1 to calculate the value for each rotation:
  • Update the currVal using the formula mentioned above.
  • Update the maximum sum accordingly in each step.
• Return the maximum value as the required answer.

Below is the implementation of the above idea.

C++

// C++ program to find max value of i*arr[i]
#include <iostream>
using namespace std;

// Returns max possible value of i*arr[i]
int maxSum(int arr[], int n)
{
    // Find array sum and i*arr[i] with no rotation
    int arrSum = 0; // Stores sum of arr[i]
    int currVal = 0; // Stores sum of i*arr[i]
    for (int i = 0; i < n; i++) {
        arrSum = arrSum + arr[i];
        currVal = currVal + (i * arr[i]);
    }

    // Initialize result as 0 rotation sum
    int maxVal = currVal;

    // Try all rotations one by one and find
    // the maximum rotation sum.
    for (int j = 1; j < n; j++) {
        currVal = currVal + arrSum - n * arr[n - j];
        if (currVal > maxVal)
            maxVal = currVal;
    }

    // Return result
    return maxVal;
}

// Driver program
int main(void)
{
    int arr[] = { 10, 1, 2, 3, 4, 5, 6, 7, 8, 9 };
    int n = sizeof(arr) / sizeof(arr[0]);
    cout << "\nMax sum is " << maxSum(arr, n);
    return 0;
}

Java

// Java program to find max value of i*arr[i]

import java.util.Arrays;

class Test {
    static int arr[]
        = new int[] { 10, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

    // Returns max possible value of i*arr[i]
    static int maxSum()
    {
        // Find array sum and i*arr[i] with no rotation
        int arrSum = 0; // Stores sum of arr[i]
        int currVal = 0; // Stores sum of i*arr[i]
        for (int i = 0; i < arr.length; i++) {
            arrSum = arrSum + arr[i];
            currVal = currVal + (i * arr[i]);
        }

        // Initialize result as 0 rotation sum
        int maxVal = currVal;

        // Try all rotations one by one and find
        // the maximum rotation sum.
        for (int j = 1; j < arr.length; j++) {
            currVal = currVal + arrSum
                      - arr.length * arr[arr.length - j];
            if (currVal > maxVal)
                maxVal = currVal;
        }

        // Return result
        return maxVal;
    }

    // Driver method to test the above function
    public static void main(String[] args)
    {
        System.out.println("Max sum is " + maxSum());
    }
}

Python

'''Python program to find maximum value of Sum(i*arr[i])'''

# returns max possible value of Sum(i*arr[i])


def maxSum(arr):

    # stores sum of arr[i]
    arrSum = 0

    # stores sum of i*arr[i]
    currVal = 0

    n = len(arr)

    for i in range(0, n):
        arrSum = arrSum + arr[i]
        currVal = currVal + (i*arr[i])

    # initialize result
    maxVal = currVal

    # try all rotations one by one and find the maximum
    # rotation sum
    for j in range(1, n):
        currVal = currVal + arrSum-n*arr[n-j]
        if currVal > maxVal:
            maxVal = currVal

    # return result
    return maxVal


# test maxsum(arr) function
if __name__ == '__main__':
    arr = [10, 1, 2, 3, 4, 5, 6, 7, 8, 9]
    print "Max sum is: ", maxSum(arr)

C#

// C# program to find max value of i*arr[i]
using System;

class Test {
    static int[] arr
        = new int[] { 10, 1, 2, 3, 4, 5, 6, 7, 8, 9 };

    // Returns max possible value of i*arr[i]
    static int maxSum()
    {
        // Find array sum and i*arr[i]
        // with no rotation
        int arrSum = 0; // Stores sum of arr[i]
        int currVal = 0; // Stores sum of i*arr[i]

        for (int i = 0; i < arr.Length; i++) {
            arrSum = arrSum + arr[i];
            currVal = currVal + (i * arr[i]);
        }

        // Initialize result as 0 rotation sum
        int maxVal = currVal;

        // Try all rotations one by one and find
        // the maximum rotation sum.
        for (int j = 1; j < arr.Length; j++) {
            currVal = currVal + arrSum
                      - arr.Length * arr[arr.Length - j];
            if (currVal > maxVal)
                maxVal = currVal;
        }

        // Return result
        return maxVal;
    }

    // Driver Code
    public static void Main()
    {
        Console.WriteLine("Max sum is " + maxSum());
    }
}

// This article is contributed by vt_m.

PHP

<?php
// PHP program to find max 
// value of i*arr[i] 

// Returns max possible 
// value of i*arr[i]
function maxSum($arr, $n)
{
    // Find array sum and
    // i*arr[i] with no rotation
    
    // Stores sum of arr[i]
    $arrSum = 0; 
    
    // Stores sum of i*arr[i]
    $currVal = 0; 
    for ($i = 0; $i < $n; $i++)
    {
        $arrSum = $arrSum + $arr[$i];
        $currVal = $currVal + 
                  ($i * $arr[$i]);
    }

    // Initialize result as
    // 0 rotation sum
    $maxVal = $currVal;

    // Try all rotations one 
    // by one and find the 
    // maximum rotation sum.
    for ($j = 1; $j < $n; $j++)
    {
        $currVal = $currVal + $arrSum - 
                   $n * $arr[$n - $j];
        if ($currVal > $maxVal)
            $maxVal = $currVal;
    }

    // Return result
    return $maxVal;
}

// Driver Code
$arr = array (10, 1, 2, 3, 4,
              5, 6, 7, 8, 9);
$n = sizeof($arr);
echo "Max sum is " ,
     maxSum($arr, $n);

// This code is contributed by m_kit
?>

Javascript

<script>
// JavaScript program to find max value of i*arr[i]

// Returns max possible value of i*arr[i]
function maxSum(arr, n)
{
    // Find array sum and i*arr[i] with no rotation
    let arrSum = 0; // Stores sum of arr[i]
    let currVal = 0; // Stores sum of i*arr[i]
    for (let i=0; i<n; i++)
    {
        arrSum = arrSum + arr[i];
        currVal = currVal+(i*arr[i]);
    }

    // Initialize result as 0 rotation sum
    let maxVal = currVal;

    // Try all rotations one by one and find
    // the maximum rotation sum.
    for (let j=1; j<n; j++)
    {
        currVal = currVal + arrSum-n*arr[n-j];
        if (currVal > maxVal)
            maxVal = currVal;
    }

    // Return result
    return maxVal;
}

// Driver program
    let arr = [10, 1, 2, 3, 4, 5, 6, 7, 8, 9];
    let n = arr.length;
    document.write("Max sum is " + maxSum(arr, n));






// This code is contributed by Surbhi Tyagi.
</script>

Output

Max sum is 330

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