标签：char end start res Substring 76 Window dict my

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

• If there is no such window in S that covers all characters in T, return the empty string "".
• If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

Time: O(N)

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        res = ''
        my_dict = {}
        for char in t:
            freq = my_dict.get(char, 0)
            my_dict[char] = freq + 1
        count, start, end, min_len = len(my_dict), 0, 0, sys.maxsize
        res_end, res_start = 0, 0
        while end != len(s):
            char = s[end]
            if char in my_dict:
                my_dict[char] -= 1
                if my_dict[char] == 0:
                    count -= 1
            end += 1
            
            while count == 0:    
                start_char = s[start]
                if start_char in my_dict:
                    my_dict[start_char] += 1
                    if my_dict[start_char] > 0:
                        count += 1            
            
                if min_len > end - start:
                    min_len = end - start
                    res_end = end
                    res_start = start
                start += 1
                
        return s[res_start: res_end]
            

标签：char,end,start,res,Substring,76,Window,dict,my
来源： https://www.cnblogs.com/xuanlu/p/11660721.html

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