76. 最小覆盖子串

题目链接	76. 最小覆盖子串
思路	滑动窗口
题解链接	两种方法：从 O(52m+n) 到 O(m+n)，附题单（Python/Java/C++/C/Go/JS/Rust）
关键点	如何跟踪子串与目标串的差距量（即less）
时间复杂度	\(O(n+m)\)
空间复杂度	\(O(|\Sigma|)\)

代码实现：

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        cnt = defaultdict(int)
        for ch in t:
            cnt[ch] += 1
        less = len(cnt)

        answer_left, answer_right = -1, len(s)
        left = 0
        for right, ch in enumerate(s):
            cnt[ch] -= 1
            if cnt[ch] == 0:
                less -= 1
            while less == 0:
                if right - left < answer_right - answer_left:
                    answer_left, answer_right = left, right
                left_ch = s[left]
                if cnt[left_ch] == 0:
                    less += 1
                cnt[left_ch] += 1
                left += 1
        return "" if answer_left < 0 else s[answer_left:answer_right+1]