[python 刷题] 76 Minimum Window Substring

题目：

Given two strings s and t of lengths m and n respectively, return the minimum window substring of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string “”.

The testcases will be generated such that the answer is unique.

这道题我用的解法和 [python 刷题] 567 Permutation in String 一样的，不过 [python 刷题] 567 Permutation in String 要求找到完全一致的配对，而这道题可以允许子字符串包含重复和多余的字符。

以 LC 给的案例来说 Input: s = "ADOBECODEBANC", t = "ABC" ，满足 ABC 只出现了一次的子字符串有三个：

题目中要求的是长度最短的字符串，因此满足要求的就是 BANC

[python 刷题] 567 Permutation in String 是在满足需求，即找到包含 t 的子字符串就返回 True，这里只需要做一个最简单的修改，就是完成对整个数组的循环，并且在满足 !current_selected_str 或 r - l + 1 < current_selected_str 两个情况下，更新 current_selected_str 即可

代码如下：

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        tc, sc = collections.Counter(t), collections.Counter()

        def has_substring():
            for c in tc:
                if sc[c] < tc[c]:
                    return False

            return True

        l = 0
        res = ''
        for r, c in enumerate(s):
            sc[c] += 1

            while sc[s[l]] > tc[s[l]] and l < r:
                sc[s[l]] -= 1
                l += 1

            if has_substring():
                if not len(res) or r - l + 1 < len(res):
                    res = s[l:r + 1]

        return res