LeetCode 552. Student Attendance Record II 斐波拉契 log(n)

最新推荐文章于 2025-01-20 07:15:00 发布

原创最新推荐文章于 2025-01-20 07:15:00 发布 · 205 阅读

0 ·

CC 4.0 BY-SA版权

算法同时被 2 个专栏收录

475 篇文章

订阅专栏

leetcode

231 篇文章

订阅专栏

Given a positive integer n, return the number of all possible attendance records with length n, which will be regarded as rewardable. The answer may be very large, return it after mod 109 + 7.

A student attendance record is a string that only contains the following three characters:

'A' : Absent.
'L' : Late.
'P' : Present.

A record is regarded as rewardable if it doesn't contain more than one 'A' (absent) or more than two continuous 'L' (late).

Example 1:

Input: n = 2
Output: 8 
Explanation:
There are 8 records with length 2 will be regarded as rewardable:
"PP" , "AP", "PA", "LP", "PL", "AL", "LA", "LL"
Only "AA" won't be regarded as rewardable owing to more than one absent times.

Note: The value of n won't exceed 100,000.

------------------------------------------------------

DP of O(n) is easy:

class Solution:
    def checkRecord(self, n: int) -> int:
        #fp[i] =fp[i-1]+fl[i-1]
        #fl[i] =fp[i-1]
        #fa[i] =fp[i-1]+fl[i-1]
        #fap[i]=fap[i-1]+fal[i-1]+fa[i-1]
        #fal[i]=fap[i-1]+fa[i-1]
        large = 10**9+7
        fp_pre,fl_pre,fa_pre,fap_pre,fal_pre,fall_pre,fll_pre = 1,1,1,0,0,0,0
        for i in range(1,n):
            fp = (fp_pre+fl_pre+fll_pre)%large
            fl = fp_pre%large
            fa = (fp_pre+fl_pre+fll_pre)%large
            fap = (fap_pre+fa_pre+fal_pre+fall_pre)%large
            fal = (fap_pre+fa_pre)%large
            fall = fal_pre%large
            fll = fl_pre%large
            fp_pre,fl_pre,fa_pre,fap_pre,fal_pre,fall_pre,fll_pre = fp,fl,fa,fap,fal,fall,fll
        print(fp_pre,fl_pre,fa_pre,fap_pre,fal_pre,fall_pre,fll_pre)
        return (fp_pre+fl_pre+fa_pre+fap_pre+fal_pre+fall_pre+fll_pre)%large

Improve it to O(log(n)) by fib intution:

import numpy as np
class Solution(object):
    # fp_pre0, fl_pre1, fa_pre2, fap_pre3, fal_pre4, fall_pre5, fll_pre6 = 1, 1, 1, 0, 0, 0, 0
    # for i in range(1, n):
    #     fp = (fp_pre + fl_pre + fll_pre) % large
    #     fl = fp_pre % large
    #     fa = (fp_pre + fl_pre + fll_pre) % large
    #     fap = (fap_pre + fa_pre + fal_pre + fall_pre) % large
    #     fal = (fap_pre + fa_pre) % large
    #     fall = fal_pre % large
    #     fll = fl_pre % large
    #
    def checkRecord(self, n):
        A = np.matrix([[1,1,0,0,0,0,1],
                       [1,0,0,0,0,0,0],
                       [1,1,0,0,0,0,1],
                       [0,0,1,1,1,1,0],
                       [0,0,1,1,0,0,0],
                       [0,0,0,0,1,0,0],
                       [0,1,0,0,0,0,0]])
        #print(A**n)
        n -= 1
        power = A
        mod = 10**9 + 7
        while n:
            if n & 1:
                power = (power * A) % mod
            A = A**2 % mod
            n //= 2
        #print(power)
        return int(sum(power[:,0]))%mod