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 = 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
So the maximum value of F(0), F(1), F(2), F(3) is F(3) = 26.
Related Topics:
Math
F(0) = (0 * A[0]) + (1 * A[1]) + ... + ((N - 1) * A[N - 1])
F(1) = (1 * A[0]) + (2 * A[1]) + ... + (0 * A[N - 1])
= F(0) + Sum(A) - N * A[N - 1]
F(2) = F(1) + Sum(A) - N * A[N - 2]
...
F(k) = F(k - 1) + Sum(A) - N * A[N - k]
So we can compute each F(k) in O(1) time and get the max in O(N) time.
// OJ: https://leetcode.com/problems/rotate-function/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
int maxRotateFunction(vector<int>& A) {
if (A.empty()) return 0;
long long f = 0, ans = INT_MIN, N = A.size(), sum = accumulate(A.begin(), A.end(), (long long)0);
for (int i = 0; i < N; ++i) f += i * A[i];
for (int i = N - 1; i >= 0; --i) ans = max(ans, f += (sum - N * A[i]));
return ans;
}
};