# Rotate Function in C++

Suppose we have Given an array of integers A and let n is the length of array A. Now assume Bk to be an array obtained by rotating the array A, k positions clock-wise. Here the rotation can be defined as −

• F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1].

Now find the maximum value of F(0), F(1), ..., F(n-1).

So if the input is like A = [4,3,2,6], then −

• F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25

• F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16

• F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23

• F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26

The maximum is 26.

To solve this, we will follow these steps −

• n := size of array A, if n is 0, then return 0

• ret := 0, create two arrays of size n, these are right and left

• left[0] := A[0]

• for i in range 1 to n – 1

• left[i] := left[i] + left[i – 1]

• left[i] := left[i] + A[i]

• right[n – 1] := A[n – 1]

• for i in range n – 1 down to 0

• right[i] := right[i] + right[i + 1]

• right[i] := right[i] + A[i]

• rightMul := 0, cnt := n – 1

• for i in range n – 1 down to 1

• rightMul := rightMul + A[i] * cnt

• decrease cnt by 1

• Make an array x of size n

• for i in range 0 to n – 2

• x[i] := rightMul

• rightMul := rightMul – right[i + 1]

• leftMul := 0, cnt := 1

• for i in range 0 down to n – 2

• leftMul := leftMul + A[i] * cnt

• increase cnt by 1

• decrease cnt by 1

• for i in range n – 1 down to 1

• x[i] := x[i] + leftMul

• leftMul := leftMul – A[i – 1] * cnt

• if i – 2 >= 0, then leftMul := leftMul + left[i – 2]

• return max of x

## Example (C++)

Let us see the following implementation to get better understanding −

Live Demo

#include <bits/stdc++.h>
using namespace std;
typedef long long int lli;
class Solution {
public:
int maxRotateFunction(vector<int>& A) {
lli n = A.size();
if(n == 0) return 0;
lli ret = 0;
vector <lli>right(n);
vector <lli> left(n);
left[0] = A[0];
for(lli i = 1; i < n; i++){
left[i] += left[i - 1];
left[i] += A[i];
}
right[n - 1] = A[n - 1];
for(lli i = n - 2; i >= 0; i--){
right[i] += right[i + 1];
right[i] += A[i];
}
lli rightMul = 0;
lli cnt = n - 1;
for(lli i = n - 1; i > 0; i--){
rightMul += (A[i] * cnt);
cnt--;
}
vector <lli> x(n);
for(lli i = 0; i < n - 1; i++){
x[i] = rightMul;
rightMul -= right[i + 1];
}
lli leftMul = 0;
cnt = 1;
for(lli i = 0; i < n - 1; i++){
leftMul += A[i] * cnt;
cnt++;
}
cnt--;
for(lli i = n - 1; i >= 1; i--){
x[i] += leftMul;
leftMul -= (A[i - 1] * cnt);
if(i - 2 >= 0) leftMul += left[i - 2];
}
ret = INT_MIN;
for(lli i = 0; i < x.size(); i++) ret = max(ret, x[i]);
return ret;
}
};
main(){
Solution ob;
vector<int> v = {4,3,2,6};
cout <<(ob.maxRotateFunction(v));
}

## Input

[4,3,2,6]

## Output

26