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

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Problem statement

You are given an array 'ARR' consisting of 'N' elements, and you need to find the maximum value of sum(i * ARR[i]) among all possible rotations of the given array. The rotations allowed are left rotation, and right rotation, and these rotations can be performed any number of times on the array.

Sum(i * ARR[i]) denotes the summation of all values (i * ARR[i]) for every 'i', where 0 <= 'i' <= 'N' - 1.

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.
Detailed explanation ( Input/output format, Notes, Images )
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
Sample Input 1 :
1
5
1 5 2 10 0
Sample Output 1 :
57
Explanation For Sample Input 1 :
Sum of all i*ARR[i] values after 'X' rotations are as follows:
After 0 rotations (original array): 0*1 + 1*5 + 2*2 + 3*10 + 4*0 = 39.
We are choosing to rotate left.
After 1 rotation to the left on the original array: 0*5 + 1*2 + 2*10 + 3*0 + 4*1 = 26.
After 2 rotations to the left on the original array: 0*2 + 1*10 + 2*0 + 3*1 + 4*5 = 33.
After 3 rotations to the left on the original array: 0*10 + 1*0 + 2*1 + 3*5 + 4*2 = 25.  
After 4 rotations to the left on the original array: 0*0 + 1*1 + 2*5 + 3*2 + 4*10 = 57.
57 is the maximum value of sum(i*ARR[i]) among all rotations of the given array.
Sample Input 2 :
1
4
1 2 3 4
Sample Output 2 :
20
Explanation For Sample Input 2 :
The original array has the maximum sum among all possible rotations of the array. The original array sums to 20, as 1*0 + 2*1 + 3*2 + 4*3 = 20.
Hint

Find maximum from all possible rotations.

Approaches (2)
Brute Force

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.
Time Complexity

O(N^2), where ‘N’ is the size of the given array.

 

We find all possible rotations for each index ‘i’ and then calculate the ‘currentSum’ by again traversing the array. Hence, the overall time complexity will be O(N^2).

Space Complexity

O(1)

 

We are not using any extra space. Hence, the overall space complexity will be O(1).

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