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

  • Last Updated : 03 Jan, 2022

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)
    • Auxiliary Space : O(1)

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)

    Method 3: The method discusses the solution using pivot in O(n) time. The pivot method can only be used in the case of a sorted or a rotated sorted array. For example: {1, 2, 3, 4} or {2, 3, 4, 1}, {3, 4, 1, 2} etc.

    • 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:

    Java




    // Java program to find maximum sum 
    // of all rotation of i*arr[i] using pivot.
      
    import java.util.*;
    import java.lang.*;
    import java.io.*;
      
    class GFG
    {
      
    // function definition 
    static int maxSum(int arr[], int n) 
    {
        int sum = 0;
        int i;
        int pivot = findPivot(arr, n);
      
        // difference in pivot and index of
        // last element of array
        int diff = n - 1 - pivot; 
        for(i = 0; i < n; i++)
        
            sum= sum + ((i + diff) % n) * arr[i];
        }
        return sum;
    }
      
    // function to find pivot
    static int findPivot(int arr[], int n)
    {
        int i;
        for(i = 0; i < n; i++)
        {
            if(arr[i] > arr[(i + 1) % n])
                return i;
        }
        return 0;
    }
      
    // Driver code
    public static void main(String args[])
    {
        // rotated input array
        int arr[] = {8, 3, 1, 2}; 
        int n = arr.length;
        int max = maxSum(arr,n); 
        System.out.println(max);
          
    }
    }

    Output: 

    29
    • 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)

    Please refer complete article on Maximum sum of i*arr[i] among all rotations of a given array for more details!


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