Rotate Array
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Given an array of integers A
and let n to be its length.
Assume Bk
to be an array obtained by rotating the array A
k positions clock-wise, we define a "rotation function" F
on A
as follow:
F(k) = 0 * Bk[0] + 1 * Bk[1] + ... + (n-1) * Bk[n-1]
.
Calculate the maximum value of F(0), F(1), ..., F(n-1)
.
Note:
n is guaranteed to be less than 105.
Example:
A = [4, 3, 2, 6]F(0) = (0 * 4) + (1 * 3) + (2 * 2) + (3 * 6) = 0 + 3 + 4 + 18 = 25F(1) = (0 * 6) + (1 * 4) + (2 * 3) + (3 * 2) = 0 + 4 + 6 + 6 = 16F(2) = (0 * 2) + (1 * 6) + (2 * 4) + (3 * 3) = 0 + 6 + 8 + 9 = 23F(3) = (0 * 3) + (1 * 2) + (2 * 6) + (3 * 4) = 0 + 2 + 12 + 12 = 26So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
According to the problem statement
F(0) = (0A) + (1B) + (2C) + (3D) + (4E)
F(1) = (4A) + (0B) + (1C) + (2D) + (3E)
F(2) = (3A) + (4B) + (0C) + (1D) + (2*E)
This problem at a glance seem like a difficult problem. I am not very strong in mathematics, so this is how I visualize this problem
We can construct F(1) from F(0) by two step:
Step 1. taking away one count of each coin from F(0), this is done by subtracting "sum" from "iteration" in the code below
after step 1 F(0) = (-1A) + (0B) + (1C) + (2D) + (3*E)
Step 2. Add n times the element which didn't contributed in F(0), which is A. This is done by adding "A[j-1]len" in the code below.
after step 2 F(0) = (4A) + (0B) + (1C) + (2D) + (3E)
At this point F(0) can be considered as F(1) and F(2) to F(4) can be constructed by repeating the above steps.
By this mind, I solve this problem using c++
class Solution {
public:
int maxRotateFunction(vector<int>& A) {
if(A.size() == 0) return 0;
int sum = 0,iterator = 0;
int len = A.size();
for(int i = 0;i < len;++i){ sum += A[i];iterator += (i * A[i]);}
int res = iterator;
for(int j = 0;j < len -1;++j){
iterator = iterator - sum + A[j]*len;
res = max(res,iterator);
}
return res;
}
};
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