Solution

Counter奇技淫巧版本（1276ms）：

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        left, right = 0, 0  # [left, right)
        need = Counter(t)
        window = Counter()
        ans = [0, len(s) + 1]

        while right < len(s):
            c = s[right]  # 1. 增大窗口，寻找可行解
            right += 1
            if c in need:
                window[c] += 1
	          
            while left < right and len(+(need - window)) == 0:
                if right - left < ans[1] - ans[0]:  # 只需要在这里更新答案！
                    ans = [left, right]

                d = s[left]  # 2. 缩小窗口，优化可行解
                left += 1
                if d in need:
                    window[d] -= 1
							  
								# 为什么不需要在这里更新呢？因为若满足条件，一定会进入内层循环，使用上面的代码更新
               
        return s[ans[0]:ans[1]] if ans[1] != len(s) + 1 else ""

正常版本（120ms）：

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        left, right = 0, 0  # [left, right)
        need = Counter(t)
        window = Counter()
        valid = 0
        ans = [0, len(s) + 1]

        while right < len(s):
            c = s[right]  # 1. 增大窗口，寻找可行解
            right += 1
            if c in need:
                window[c] += 1
                if window[c] == need[c]:
                    valid += 1
	          
            while left < right and valid == len(need):
                if right - left < ans[1] - ans[0]:  # 只需要在这里更新答案！
                    ans = [left, right]

                d = s[left]  # 2. 缩小窗口，优化可行解
                left += 1
                if d in need:
                    if window[d] == need[d]:
                        valid -= 1
                    window[d] -= 1

        return s[ans[0]:ans[1]] if ans[1] != len(s) + 1 else ""