Minimum Window Substring

76. Minimum Window Substring

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.

Example 1:

Input: s = "ADOBECODEBANC", t = "ABC"
Output: "BANC"
Explanation: The minimum window substring "BANC" includes 'A', 'B', and 'C' from string t.

Example 2:

Input: s = "a", t = "a"
Output: "a"
Explanation: The entire string s is the minimum window.

Example 3:

Input: s = "a", t = "aa"
Output: ""
Explanation: Both 'a's from t must be included in the window.
Since the largest window of s only has one 'a', return empty string.

Constraints:

• m == s.length

• n == t.length

• 1 <= m, n <= 105

• s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

分析：

双指针+counter

注意required_char=len(counter_t) not len(t) 需要unique number of char

```python3
from collections import Counter


class Solution:
    def minWindow(self, s: str, t: str) -> str:
        if len(s) < len(t):
            return ""
        min_start,min_len,left = 0, float("inf"),0
        
        left = 0
        count_window = Counter()
        count_t = Counter(t)
        formed_char = 0
        required_char = len(count_t)
        for right, char in enumerate(s):
            count_window[char] += 1
            if count_window[char] == count_t[char]:
                formed_char += 1
            while formed_char == required_char:
                if min_len > right - left + 1:
                    min_len = right - left + 1
                    min_start = left
                
                if count_window[s[left]] == count_t[s[left]]:
                    formed_char -= 1 
                count_window[s[left]] -= 1
                left += 1
        return s[min_start: min_start + min_len] if min_len != float("inf") else ""
                






```