Given an array, only rotation operation is allowed on array. We can rotate the array as many times as we want. Return the maximum possible summation of i*arr[i].
Examples :
Input: arr[] = {1, 20, 2, 10}
Output: 72
We can get 72 by rotating array twice.
{2, 10, 1, 20}
20*3 + 1*2 + 10*1 + 2*0 = 72
Input: arr[] = {10, 1, 2, 3, 4, 5, 6, 7, 8, 9}
Output: 330
We can get 330 by rotating array 9 times.
{1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
0*1 + 1*2 + 2*3 ... 9*10 = 330
We strongly recommend you to minimize your browser and try this yourself first.
A Simple Solution is to find all rotations one by one, check sum of every rotation and return the maximum sum. Time complexity of this solution is O(n2).
We can solve this problem in O(n) time using an Efficient Solution.
Let Rj be value of i*arr[i] with j rotations. The idea is to calculate next rotation value from previous rotation, i.e., calculate Rj from Rj-1. We can calculate initial value of result as R0, then keep calculating next rotation values.
How to efficiently calculate Rj from Rj-1?
This can be done in O(1) time. Below are details.
Let us calculate initial value of i*arr[i] with no rotation
R0 = 0*arr[0] + 1*arr[1] +...+ (n-1)*arr[n-1]
After 1 rotation arr[n-1], becomes first element of array,
arr[0] becomes second element, arr[1] becomes third element
and so on.
R1 = 0*arr[n-1] + 1*arr[0] +...+ (n-1)*arr[n-2]
R1 - R0 = arr[0] + arr[1] + ... + arr[n-2] - (n-1)*arr[n-1]
After 2 rotations arr[n-2], becomes first element of array,
arr[n-1] becomes second element, arr[0] becomes third element
and so on.
R2 = 0*arr[n-2] + 1*arr[n-1] +...+ (n-1)*arr[n-3]
R2 - R1 = arr[0] + arr[1] + ... + arr[n-3] - (n-1)*arr[n-2] + arr[n-1]
If we take a closer look at above values, we can observe
below pattern
Rj - Rj-1 = arrSum - n * arr[n-j]
Where arrSum is sum of all array elements, i.e.,
arrSum = ∑ arr[i]
0<=i<=n-1
Below is complete algorithm:
1) Compute sum of all array elements. Let this sum be 'arrSum'.
2) Compute R0 by doing i*arr[i] for given array.
Let this value be currVal.
3) Initialize result: maxVal = currVal // maxVal is result.
// This loop computes Rj from Rj-1
4) Do following for j = 1 to n-1
......a) currVal = currVal + arrSum-n*arr[n-j];
......b) If (currVal > maxVal)
maxVal = currVal
5) Return maxVal
Below is the implementation of above idea :
Java
// Java program to find max value of i*arr[i] import java.util.Arrays; class Test{ static int arr[] = new int[]{10, 1, 2, 3, 4, 5, 6, 7, 8, 9}; // Returns max possible value of i*arr[i] static int maxSum() { // Find array sum and i*arr[i] with no rotation int arrSum = 0; // Stores sum of arr[i] int currVal = 0; // Stores sum of i*arr[i] for (int i=0; i<arr.length; i++) { arrSum = arrSum + arr[i]; currVal = currVal+(i*arr[i]); } // Initialize result as 0 rotation sum int maxVal = currVal; // Try all rotations one by one and find // the maximum rotation sum. for (int j=1; j<arr.length; j++) { currVal = currVal + arrSum-arr.length*arr[arr.length-j]; if (currVal > maxVal) maxVal = currVal; } // Return result return maxVal; } // Driver method to test the above function public static void main(String[] args) { System.out.println("Max sum is " + maxSum()); }} |
Output :
Max sum is 330
Time Complexity : O(n)
Auxiliary Space : O(1)
Please refer complete article on Find maximum value of Sum( i*arr[i]) with only rotations on given array allowed for more details!
