Maximum sum of i*arr[i] among all rotations of a given array

Given an integer array arr[] of size n, find the maximum value of the expression i * arr[i] (for all i from 0 to n-1) after rotating the array any number of times.

Examples:

Input: arr[] = [8, 3, 1, 2]
Output: 29
Explanation: Out of all the possible configurations by rotating the elements: arr[] = [3, 1, 2, 8] here (3*0) + (1*1) + (2*2) + (8*3) sum is maximum i.e. 29.

Input: arr[] = [1, 2, 3]
Output: 8
Explanation: Out of all the possible configurations by rotating the elements: arr[] = [1, 2, 3] here (1*0) + (2*1) + (3*2) sum is maximum i.e. 8.

[Naive Approach] Using Nested Loops - O(n ^ 2) Time and O(1) Space

The idea is to find the sum of all the elements of the array for each possible rotation, and store the maximum of them as answer.

• To do so, use nested loops, where the outer loop marks the starting point of the array, and the inner loop iterates through each of the element, and store the sum of i * arr[i].
• Check if the maximum sum is greater than the current sum then update the maximum sum.

29

[Expected Approach] Using Mathematics - O(n) Time and O(1) Space

The main idea is to Instead of recalculating the weighted sum from scratch for each rotation, it first computes the total sum of the array and the initial rotation value. Then, for each subsequent rotation, it derives the new value using a formula based on the previous value and the total sum. and find the maximum value of the sum across all rotations of the array.

We can calculate the value of the next rotation using the value of the current one. Let curSum be the total sum of all elements in the array, and currVal be the value of the current rotation. Then, the value of the next rotation can be derived from currVal by adjusting it based on the shift in element positions.

finding NextVal Using currVal:

We can compute the value of the i^th rotation using the value of the (i-1)^th rotation:

Nextval = currVal - (curSum - arr[i-1]) + arr[i-1] * (n-1)

29

[Alternate Approach] Using Pivot - O(n) Time and O(1) Space

Note: This approach works only for sorted or rotated sorted arrays.

We know for an array the maximum sum will be when the array is sorted in ascending order. In case of a sorted rotated array, we can rotate the array to make it in ascending order. So, in this case, the pivot element is needed to be found following which the maximum sum can be calculated.

• Find the pivot of the array: if arr[i] > arr[(i+1)%n] then it is the pivot element. (i+1)%n is used to check for the last and first element.
• After getting pivot the sum can be calculated by finding the difference with the pivot which will be the multiplier and multiply it with the current element while calculating the sum

29