Minimum Window With Characters

Given two strings s and t, return the shortest substring of s such that every character in t, including duplicates, is present in the substring. If such a substring does not exist, return an empty string “”.

You may assume that the correct output is always unique.

Example 1:

Input: s = "OUZODYXAZV", t = "XYZ"Output: "YXAZ"

Explanation: “YXAZ” is the shortest substring that includes “X”, “Y”, and “Z” from string t.

Example 2:

Input: s = "xyz", t = "xyz"Output: "xyz"

Example 3:

Input: s = "x", t = "xy"Output: ""

Constraints:

1 <= s.length <= 1000
1 <= t.length <= 1000

s and t consist of uppercase and lowercase English letters.

Solution

It is apparently a sliding window question and we can easily count the number of characters in the window. So, the key problem becomes to check whether the current substring satisfies the requirement.

In fact, we will only add or remove on character at once, which means there will only be at most 1 kind of character satisfies or dissatisfies the requirements. Therefore, we can maintain the number of requirements that has been satisfied and change this number will adding/removing characters.

Code

class Solution:def minWindow(self, s: str, t: str) -> str:t_stat = {}win_stat = {}for c in range(ord('A'), ord('Z')+1):t_stat[chr(c)] = 0win_stat[chr(c)] = 0for c in range(ord('a'), ord('z')+1):t_stat[chr(c)] = 0win_stat[chr(c)] = 0satis_num = 0for c in t:t_stat[c] += 1for key in t_stat.keys():if t_stat[key] <= win_stat[key]:satis_num += 1le = 0ans_le, ans_ri = -1001, -1for ri in range(len(s)):win_stat[s[ri]] += 1if win_stat[s[ri]] == t_stat[s[ri]]:satis_num += 1print(satis_num)print(le, ri)while le <= ri and satis_num >= 52:if ri-le+1 < ans_ri-ans_le+1:ans_le = leans_ri = riif win_stat[s[le]] == t_stat[s[le]]:satis_num -= 1win_stat[s[le]] -= 1le += 1if ans_le < 0:return ''return s[ans_le: ans_ri+1]