LeetCode Minimum Window Substring

LeetCode Minimum Window Substring
LeetCode-solving-Minimum Window Substring

Original question

Given two strings S and T, it is required to find the shortest substring of S containing all characters in T in O (n) time.

Note:

If no sub-string meets the requirements, "" is returned. If multiple sub-strings meet the requirements, only one of them is possible.

Example:

Input: s = "ADOBECODEBANC", t = "ABC"

Output: "BANC"

Solutions

Determine the current sub-string through the front and back pointers. Move the sub-string until the sub-string contains all the characters in T and try to move the front pointer back, and refresh the shortest length and the corresponding starting position. If the pre-pointer does not contain any characters in T, move the pre-pointer. Move the front and back pointers alternately until the complete string S is traversed.

AC Source Code

from collections import defaultdictclass Solution(object):    def minWindow(self, s, t):        """        :type s: str        :type t: str        :rtype: str        """        MAX_INT = 2147483647        start = end = 0        char_need = defaultdict(int)    # the count of char needed by current window, negative means current window has it but not needs it        count_need = len(t)             # count of chars not in current window but in t        min_length = MAX_INT        min_start = 0        for i in t:            # current window needs all char in t            char_need[i] += 1                   while end < len(s):            if char_need[s[end]] > 0:                count_need -= 1            # current window contains s[end] now, so does not need it any more            char_need[s[end]] -= 1                  end += 1            while count_need == 0:                if min_length > end - start:                    min_length = end - start                    min_start = start                # current window does not contain s[start] any more                char_need[s[start]] += 1                    # when some count in char_need is positive, it means there is char in t but not current window                if char_need[s[start]] > 0:                     count_need += 1                start += 1        return "" if min_length == MAX_INT else s[min_start:min_start + min_length]if __name__ == "__main__":    assert Solution().minWindow("ADOBECODEBANC", "ABC") == "BANC"