How to efficiently calculate Rj from Rj-1?

Software Development
#1

This can be done in O(1) time. Below are details.

Let us calculate initial value of iarr[i] with no rotation R0 = 0arr[0] + 1arr[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 = 0arr[n-1] + 1arr[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 = 0arr[n-2] + 1arr[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 iarr[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-narr[n-j]; …b) If (currVal > maxVal) maxVal = currVal 5) Return maxVal

Below is the implementation of above idea :

  • C++

// C++ program to find max value of i*arr[i]

#include <iostream>

using namespace std;

// Returns max possible value of i*arr[i]

int maxSum( int arr[], int n)

{

`` // 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<n; 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<n; j++)

`` {

`` currVal = currVal + arrSum-n*arr[n-j];

`` if (currVal > maxVal)

`` maxVal = currVal;

`` }

`` // Return result

`` return maxVal;

}

// Driver program

int main( void )

{

`` int arr[] = {10, 1, 2, 3, 4, 5, 6, 7, 8, 9};

`` int n = sizeof (arr)/ sizeof (arr[0]);

`` cout << "\nMax sum is " << maxSum(arr, n);

`` return 0;

}

Output :

Max sum is 330

Time Complexity : O(n)
Auxiliary Space : O(1)