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

Last Updated : 28 Dec, 2021

Given an array, only rotation operation is allowed on array. We can rotate the array as many times as we want. Return the maximum possible summation of i*arr[i].

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

We strongly recommend you to minimize your browser and try this yourself first.
A Simple Solution is to find all rotations one by one, check sum of every rotation and return the maximum sum. Time complexity of this solution is O(n²).

We can solve this problem in O(n) time using an Efficient Solution.
Let R_j be value of i*arr[i] with j rotations. The idea is to calculate next rotation value from previous rotation, i.e., calculate R_j from R_j-1. We can calculate initial value of result as R₀, then keep calculating next rotation values.

How to efficiently calculate R_j from R_j-1?
This can be done in O(1) time. Below are 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

Below is complete algorithm:

1) Compute sum of all array elements. Let this sum be 'arrSum'.

2) Compute R₀ by doing i*arr[i] for given array. 
   Let this value be currVal.

3) Initialize result: maxVal = currVal // maxVal is result.

// This loop computes R_j from  R_j-1 
4) Do following for j = 1 to n-1
......a) currVal = currVal + arrSum-n*arr[n-j];
......b) If (currVal > maxVal)
            maxVal = currVal   

5) Return maxVal

Below is the implementation of above idea :

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());
    }
}