# 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.

A substring is a contiguous sequence of characters within the string.

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.
 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 `````` ``````class Solution: def minWindow(self, s: str, t: str) -> str: need = collections.Counter(t) missing = len(t) start, end = 0, 0 i = 0 for j, c in enumerate(s, 1): # index j from 1 if need[c] > 0: # included a char in t missing -= 1 need[c] -= 1 if missing == 0: # match all chars, advance left pointer while i < j and need[s[i]] < 0: need[s[i]] += 1 i += 1 # make sure the first appearing char satisfies need[char]>0 need[s[i]] += 1 # we missed this first char, so add missing by 1 missing += 1 if end == 0 or j-i < end-start: start, end = i, j i += 1 return s[start:end]``````