# Maximum Sum Of (i * ARR[i]) Among All Possible Rotations Of An Array

Last Updated: 26 Nov, 2020
Difficulty: Moderate

## PROBLEM STATEMENT

#### Note :

``````1. The array follows 0-based indexing.
2. In one rotation operation, all elements of the array will shift either towards the left or right by one index.
3. The element at the extreme left index (i.e. 0th index) will shift to the 'N-1'th index after applying one left rotation on the array and the rest all the elements will shift to the 'i-1'th index.
4. The element at the extreme right index (i.e. 'N-1'th index) will shift to the 0th index after applying one right rotation on the array and the rest of the elements will shift to the 'i' + 1'th index.
``````
##### Input Format :
``````The first line of the input contains an integer 'T' denoting the number of test cases.

The first line of each test case contains the integer 'N', denoting the size of the array.

The second line of each test case contains 'N' space-separated integers denoting the array elements.
``````
##### Output Format :
``````The only line of output of each test case should contain an integer, denoting the maximum value of sum(i * ARR[i]).
``````
##### Note :
``````You don't need to print anything. It has already been taken care of. Just implement the given function.
``````
##### Constraints :
``````1 <= T <= 10^2
1 <= N <= 10^4
0 <= ARR[i] <= 10^6

Time Limit : 1sec
``````

## Approach 1

It’s important to note that, there are only a finite number of unique arrangments of the array possible after applying rotation operations on the array any number of times. So what we can do is find the sum(i*ARR[i]) for all possible rotations and return the maximum among them.

Here is the algorithm :

1. Create a variable (say, ‘maxSum’) and initialise it to 0.
2. Run a loop from 0 to 'N - 1' (say, iterator ‘i’), as these are all possible starting indices of the array in some rotation or the other.
3. For each starting index, calculate the ending index of the array possible as ‘endIndex’ =  (i + N - 1) % N (because the size of the array is ‘N’) and store the current array sum in variable (say, ‘currentSum’)
4. Iterate from starting index to next 'N - 1'(valid) indices by calculating the next index as (i + 1) % N till you reach end index and update the ‘currentSum’ by adding 'i * ARR[i]'.
5. After reaching the end index update ‘maxSum’ to maximum of 'maxSum' and ‘currentSum’.
6. Finally, return the maximum sum after all iterations of i.