424. Longest Repeating Character Replacement

Problem

You are given a string s and an integer k. You can choose any character of the string and change it to any other uppercase English character. You can perform this operation at most k times.

Return the length of the longest substring containing the same letter you can get after performing the above operations.

https://leetcode.com/problems/longest-repeating-character-replacement/

Example 1:

Input: s = "ABAB", k = 2
Output: 4
Explanation: Replace the two 'A's with two 'B's or vice versa.

Example 2:

Input: s = "AABABBA", k = 1
Output: 4
Explanation: Replace the one 'A' in the middle with 'B' and form "AABBBBA".
The substring "BBBB" has the longest repeating letters, which is 4.
There may exists other ways to achieve this answer too.

Constraints:

1 <= s.length <= 10^5
s consists of only uppercase English letters.
0 <= k <= s.length

Test Cases

1 2	class Solution: def characterReplacement(self, s: str, k: int) -> int:

solution_test.py

import pytest

from solution import Solution


@pytest.mark.parametrize('s, k, expected', [
    ('ABAB', 2, 4),
    ('AABABBA', 1, 4),
])
class Test:
    def test_solution(self, s, k, expected):
        sol = Solution()
        assert sol.characterReplacement(s, k) == expected

Thoughts

先考虑某一个字母，比如 ‘A’。最多把 k 个其他字母改成 ‘A’，看能得到的最长的连续 ‘A’ 有多长。

设当前的子串 s[i:j] 是连续的 ‘A’，且其中有 r 个 ‘A’ 是从其他字母换过来的（0 <= i < j <= n，0 <= r <= k）。

如果 s[j] 是 ‘A’，或者还有替换次数尚未用完，则可以右移 j（并更新 k 的使用次数）。

否则根据 s[i] 释放 k 的使用次数，并向右移动 i。

设各不相同的字母总数是 C，那么时间复杂度是 O(C * n)，空间复杂度是 O(C)。

PS：之前用这种滑窗法的时候，写循环的时候总是被边界条件搞得头疼。主要是在想思路的时候，往往会让区间的一头直接移动到最远，然后再移动另一头。但如果按这个方式写代码，就会引入嵌套的循环，增加更多的边界条件判定，代码看着也复杂。实际上代码里应该就一个循环，每次循环只移动一步。比如即使区间的右边界可以连续移动五步，也不要在主循环体内直接让它移动五步，而是要正常地循环五次，每次都移动一步。

Code

solution.py

class Solution:
    def characterReplacement(self, s: str, k: int) -> int:
        n = len(s)
        chars = set(s)
        max_len = 0

        for c in chars:
            i = 0
            j = 0
            used = 0
            while j < n:
                if s[j] == c:
                    j += 1
                elif used < k:
                    used += 1
                    j += 1
                else:
                    if s[i] != c:
                        used -= 1
                    i += 1

                max_len = max(max_len, j - i)

        return max_len

似乎不是很快，先这样吧。

本文除了题目描述（Problem）部分之外，其他内容由 Calf 原创，采用 CC BY-NC-SA 4.0 许可协议，转载请注明出处。