Simple Python sliding window solution with detailed explanation
\nfrom collections import Counter\n\nclass Solution:\n    def minWindow(self, s: str, t: str) -> str:\n        \'\'\'\n        Keep t_counter of char counts in 
```\nfrom collections import Counter\n\nclass Solution:\n    def minWindow(self, s: str, t: str) -> str:\n        \'\'\'\n        Keep t_counter of char counts in t\n        \n        We make a sliding window across s, tracking the char counts in s_counter\n        We keep track of matches, the number of chars with matching counts in s_counter and t_counter\n        Increment or decrement matches based on how the sliding window changes\n        When matches == len(t_counter.keys()), we have a valid window. Update the answer accordingly\n        \n        How we slide the window:\n        Extend when matches < chars, because we can only get a valid window by adding more.\n        Contract when matches == chars, because we could possibly do better than the current window.\n        \n        How we update matches:\n        This only applies if t_counter[x] > 0.\n        If s_counter[x] is increased to match t_counter[x], matches += 1\n        If s_counter[x] is increased to be more than t_counter[x], do nothing\n        If s_counter[x] is decreased to be t_counter[x] - 1, matches -= 1\n        If s_counter[x] is decreased to be less than t_counter[x] - 1, do nothing\n        \n        Analysis:\n        O(s + t) time: O(t) to build t_counter, then O(s) to move our sliding window across s. Each index is only visited twice.\n        O(s + t) space: O(t) space for t_counter and O(s) space for s_counter\n        \'\'\'\n        \n        if not s or not t or len(s) < len(t):\n            return \'\'\n        \n        t_counter = Counter(t)\n        chars = len(t_counter.keys())\n        \n        s_counter = Counter()\n        matches = 0\n        \n        answer = \'\'\n        \n        i = 0\n        j = -1 # make j = -1 to start, so we can move it forward and put s[0] in s_counter in the extend phase \n        \n        while i < len(s):\n            \n            # extend\n            if matches < chars:\n                \n                # since we don\'t have enough matches and j is at the end of the string, we have no way to increase matches\n                if j == len(s) - 1:\n                    return answer\n                \n                j += 1\n                s_counter[s[j]] += 1\n                if t_counter[s[j]] > 0 and s_counter[s[j]] == t_counter[s[j]]:\n                    matches += 1\n\n            # contract\n            else:\n                s_counter[s[i]] -= 1\n                if t_counter[s[i]] > 0 and s_counter[s[i]] == t_counter[s[i]] - 1:\n                    matches -= 1\n                i += 1\n                \n            # update answer\n            if matches == chars:\n                if not answer:\n                    answer = s[i:j+1]\n                elif (j - i + 1) < len(answer):\n                    answer = s[i:j+1]\n        \n        return answer\n```
['Sliding Window', 'Python', 'Python3']