leetcode 396. Rotate Function

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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 10^5.

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.

  • my solution:
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class Solution(object):
    def maxRotateFunction(self, A):
        if len(A) == 0:
            return 0
        n = len(A)
        def f(A,k):
            B = (A+A)[k:k+n]
            v = 0
            for i in range(n):
                v += i*B[i]
            return v
        return max(f(A,i) for i in range(n))

this solution is o(n^2), cost 14s

  • clear solution:
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class Solution:
    def maxRotateFunction(self, A):
        n = len(A)
        v = 0
        sumA = 0
        for i in range(n):
            v += i * A[i]
            sumA += A[i]
        max_v = v
        for i in range(n - 1, 0, -1):
            v = v + sumA - n * A[i]
            max_v = max(max_v, v)
        return max_v

this solution cost 0.02s

because f(k+1)-f(k) = nA[k] - sumA

f(k+1) = 0*A[k+1] + A[k+2] + 2*A[k+3]+…+(n-1)A[k+1+n-1]

A[k+n] = A[k]

f(k+1) = (n-1)*A[k]+0*A[k+1] + A[k+2] + 2*A[k+3]+…+(n-2)A[k+n-1]

f(k) = 0*A[k] + A[k+1] + 2*A[k+2]+…+(n-1)A[k+n-1]

so f(k+1)-f(k) = n*A[k] - (A[k] + A[k+1] +…+A[k+n-1])

= n*A[k] - sumA