Problem:
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.
Solution 1: Brute Force
Time complexity: O(n^2) TLE
Space complexity: O(1)
Solution 2: Math
F(K) = 0A[K] + 1A[K+1] + 2A[K+2] + ... + (n-2)A[K-2] + (n-1)A[K-1]
F(K-1) = 0A[K-1] + 1A[K] + 2A[K+1] + 3A[K+2] + ... + (n-1)A[K-2]
F(K) - F(K-1) = (n-1)A[K-1] - 1A[K] - 1A[K+1] - 1A[K+2] - ... - 1A[K-2]
= (n-1)A[K-1] - (1A[K] + 1A[K+1] + ... + 1A[K-2] + 1A[K-1] - 1A[K-1])
= nA[K-1] - sum(A)
compute F(0)
F(i) = F(i-1) + nA[i-1] - sum(A)
Time complexity: O(n)
Space complexity: O(1)
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// Author: Huahua // Running time: 11 ms class Solution { public: int maxRotateFunction(vector<int>& A) { const int n = A.size(); int sum = 0; int f = 0; for (int i = 0; i < n; ++i) { sum += A[i]; f += i * A[i]; } int ans = f; for (int i = 0; i < n - 1; ++i) { f = f + sum - n * A[n - i - 1]; ans = max(ans, f); } return ans; } }; |
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