LeetCode 第417场周赛个人题解

class Solution:
    def kthCharacter(self, k: int) -> str:
        s = [0]
        for i in range(k):
            ns = list((c + 1) % 26 for c in s)
            s += ns
            if len(s) > k: break
        return str(chr(s[k - 1] + ord('a')))

Q2. 元音辅音字符串计数 I

原题链接

Q2. 元音辅音字符串计数 I

思路分析

见T3

AC代码

python 复制代码

class Solution:
    def countOfSubstrings(self, s: str, k: int) -> int:
        n = len(s)
        st = set("aeiou")
        acc = [[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ]

        acc1 = [0] * (n + 1) + [n]
        for i, c in enumerate(s):
            acc[0][i + 1] = acc[0][i] + (1 if c == 'a' else 0)
            acc[1][i + 1] = acc[1][i] + (1 if c == 'e' else 0)
            acc[2][i + 1] = acc[2][i] + (1 if c == 'i' else 0)
            acc[3][i + 1] = acc[3][i] + (1 if c == 'o' else 0)
            acc[4][i + 1] = acc[4][i] + (1 if c == 'u' else 0)
            acc1[i + 1] = acc1[i] + (0 if c in st else 1)
        res = 0
        for i in range(n):
            l = bisect_left(acc1, k + acc1[i])
            r = bisect_right(acc1, k + acc1[i]) - 1
            if l > n:
                break
            
            b = max(bisect_right(acc[c], acc[c][i]) for c in range(5))

            if b > n:
                break
            l = max(i, b, l)
            if r >= l:
                res += r - l + 1
        return res

Q3. 元音辅音字符串计数 II

原题链接

Q3. 元音辅音字符串计数 II

思路分析

前缀和+二分

滑窗可以写，但是写红温了，还是写了沟槽的二分

开5个元音前缀和数组，一个辅音前缀和数组

枚举 i ，二分合法右端点，累计贡献

时间复杂度：O(nlogn)

AC代码

python 复制代码

class Solution:
    def countOfSubstrings(self, s: str, k: int) -> int:
        n = len(s)
        st = set("aeiou")
        acc = [[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ,[0] * (n + 1) + [n]
        ]

        acc1 = [0] * (n + 1) + [n]
        for i, c in enumerate(s):
            acc[0][i + 1] = acc[0][i] + (1 if c == 'a' else 0)
            acc[1][i + 1] = acc[1][i] + (1 if c == 'e' else 0)
            acc[2][i + 1] = acc[2][i] + (1 if c == 'i' else 0)
            acc[3][i + 1] = acc[3][i] + (1 if c == 'o' else 0)
            acc[4][i + 1] = acc[4][i] + (1 if c == 'u' else 0)
            acc1[i + 1] = acc1[i] + (0 if c in st else 1)
        res = 0
        for i in range(n):
            l = bisect_left(acc1, k + acc1[i])
            r = bisect_right(acc1, k + acc1[i]) - 1
            if l > n:
                break
            
            b = max(bisect_right(acc[c], acc[c][i]) for c in range(5))

            if b > n:
                break
            l = max(i, b, l)
            if r >= l:
                res += r - l + 1
        return res

Q4. 找出第 K 个字符 II

原题链接

Q4. 找出第 K 个字符 II

思路分析

递归

如上图所示，最终的数组其实就是不断二倍得到的

我们可以建立简单的递推关系

给定k，我们可以算出k是第几次操作得到的（取log2即可）

那么根据操作类型可以推出和对称位置的颜色关系（+1还是相同）

直接递归即可

时间复杂度：O(log^2 k)

AC代码

python 复制代码

import math

class Solution:
    def kthCharacter(self, k: int, operations: list[int]) -> str:
        def dfs(i: int) -> int:
            if i <= 0: return 0
            if i == 1:
                return 0
            b = math.ceil(math.log2(i)) - 1
            return (dfs(i - pow(2, b)) + (1 if operations[b] == 1 else 0)) % 26
            
        return chr(dfs(k) + ord('a'))