 GeeksforGeeks App
Open App Browser
Continue

# Java Program for Maximum sum of i*arr[i] among all rotations of a given array

Given an array arr[] of n integers, find the maximum that maximizes the sum of the value of i*arr[i] where i varies from 0 to n-1.

Examples:

```Input: arr[] = {8, 3, 1, 2}
Output: 29
Explanation: Lets look at all the rotations,
{8, 3, 1, 2} = 8*0 + 3*1 + 1*2 + 2*3 = 11
{3, 1, 2, 8} = 3*0 + 1*1 + 2*2 + 8*3 = 29
{1, 2, 8, 3} = 1*0 + 2*1 + 8*2 + 3*3 = 27
{2, 8, 3, 1} = 2*0 + 8*1 + 3*2 + 1*3 = 17

Input: arr[] = {3, 2, 1}
Output: 7
Explanation: Lets look at all the rotations,
{3, 2, 1} = 3*0 + 2*1 + 1*2 = 4
{2, 1, 3} = 2*0 + 1*1 + 3*2 = 7
{1, 3, 2} = 1*0 + 3*1 + 2*2 = 7```

Method 1: This method discusses the Naive Solution which takes O(n2) amount of time.
The solution involves finding the sum of all the elements of the array in each rotation and then deciding the maximum summation value.

• Approach:A simple solution is to try all possible rotations. Compute sum of i*arr[i] for every rotation and return maximum sum.
• Algorithm:
1. Rotate the array for all values from 0 to n.
2. Calculate the sum for each rotations.
3. Check if the maximum sum is greater than the current sum then update the maximum sum.
• Implementation:

## Java

 `// A Naive Java program to find``// maximum sum rotation``import` `java.util.*;``import` `java.io.*;` `class` `GFG {` `// Returns maximum value of i*arr[i]``static` `int` `maxSum(``int` `arr[], ``int` `n)``{``// Initialize result``int` `res = Integer.MIN_VALUE;` `// Consider rotation beginning with i``// for all possible values of i.``for` `(``int` `i = ``0``; i < n; i++)``{` `    ``// Initialize sum of current rotation``    ``int` `curr_sum = ``0``;` `    ``// Compute sum of all values. We don't``    ``// actually rotation the array, but compute``    ``// sum by finding ndexes when arr[i] is``    ``// first element``    ``for` `(``int` `j = ``0``; j < n; j++)``    ``{``        ``int` `index = (i + j) % n;``        ``curr_sum += j * arr[index];``    ``}` `    ``// Update result if required``    ``res = Math.max(res, curr_sum);``}` `return` `res;``}` `// Driver code``public` `static` `void` `main(String args[])``{``        ``int` `arr[] = {``8``, ``3``, ``1``, ``2``};``        ``int` `n = arr.length;``        ``System.out.println(maxSum(arr, n));``}` `    ` `}` `// This code is contributed by Sahil_Bansall`

Output :

`29`
• Complexity Analysis:

Time Complexity : O(n2), as we are using nested loops.
Auxiliary Space : O(1), as we are not using any extra space.

Method 2: This method discusses the efficient solution which solves the problem in O(n) time. In the naive solution, the values were calculated for every rotation. So if that can be done in constant time then the complexity will decrease.

• Approach: The basic approach is to calculate the sum of new rotation from the previous rotations. This brings up a similarity where only the multipliers of first and last element change drastically and the multiplier of every other element increases or decreases by 1. So in this way, the sum of next rotation can be calculated from the sum of present rotation.
• Algorithm:
The idea is to compute the value of a rotation using values of previous rotation. When an array is rotated by one, following changes happen in sum of i*arr[i].
1. Multiplier of arr[i-1] changes from 0 to n-1, i.e., arr[i-1] * (n-1) is added to current value.
2. Multipliers of other terms is decremented by 1. i.e., (cum_sum – arr[i-1]) is subtracted from current value where cum_sum is sum of all numbers.
```next_val = curr_val - (cum_sum - arr[i-1]) + arr[i-1] * (n-1);

next_val = Value of ∑i*arr[i] after one rotation.
curr_val = Current value of ∑i*arr[i]
cum_sum = Sum of all array elements, i.e., ∑arr[i].

Lets take example {1, 2, 3}. Current value is 1*0+2*1+3*2
= 8. Shifting it by one will make it {2, 3, 1} and next value
will be 8 - (6 - 1) + 1*2 = 5 which is same as 2*0 + 3*1 + 1*2```
• Implementation:

## Java

 `// An efficient Java program to compute``// maximum sum of i*arr[i]``import` `java.io.*;` `class` `GFG {``    ` `    ``static` `int` `maxSum(``int` `arr[], ``int` `n)``    ``{``        ``// Compute sum of all array elements``        ``int` `cum_sum = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``cum_sum += arr[i];` `        ``// Compute sum of i*arr[i] for``        ``// initial configuration.``        ``int` `curr_val = ``0``;``        ``for` `(``int` `i = ``0``; i < n; i++)``            ``curr_val += i * arr[i];` `        ``// Initialize result``        ``int` `res = curr_val;` `        ``// Compute values for other iterations``        ``for` `(``int` `i = ``1``; i < n; i++)``        ``{``            ``// Compute next value using previous``            ``// value in O(1) time``            ``int` `next_val = curr_val - (cum_sum -``                          ``arr[i-``1``]) + arr[i-``1``] *``                          ``(n-``1``);` `            ``// Update current value``            ``curr_val = next_val;` `            ``// Update result if required``            ``res = Math.max(res, next_val);``        ``}` `        ``return` `res;``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = {``8``, ``3``, ``1``, ``2``};``        ``int` `n = arr.length;``        ``System.out.println(maxSum(arr, n));``    ``}``}``// This code is contributed by Prerna Saini`

Output:

`29`
• Complexity analysis:
• Time Complexity: O(n).
Since one loop is needed from 0 to n to check all rotations and the sum of the present rotation is calculated from the previous rotations in O(1) time).
• Auxiliary Space: O(1).
As no extra space is required to so the space complexity will be O(1)
• Approach: Let’s assume the case of a sorted array. As 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.
• Algorithm:
1. 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.
2. 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
• Implementations:
• Complexity analysis:
• Time Complexity : O(n)
As only one loop was needed to traverse from 0 to n to find the pivot. To find the sum another loop was needed, so the complexity remains O(n).
• Auxiliary Space : O(1).
We do not require extra space to so the Auxiliary space is O(1)

My Personal Notes arrow_drop_up