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.

Difficulty: interview

Starter code:

class Solution:
    def minWindow(self, s: str, t: str) -> str:
        

Solutions:
class Solution:
     def minWindow(self, s, t):
         """
         :type s: str
         :type t: str
         :rtype: str
         """
         left=-1
         right = 0
         result = ""
         totalMatch = 0
         d = {}
         for c in t:
             d[c] = d.get(c, 0) + 1
 
         for right in range(len(s)):
             c = s[right]
             d[c] = d.get(c, 0) - 1
 
             # good match
             if d[c] >=0:
                 totalMatch +=1
 
                 #over match 可以不移动left
                 #total match, need to advance left
                 if totalMatch == len(t):
                     totalMatch -= 1
 
                     left +=1
                     while d[s[left]]<0:
                         d[s[left]] += 1
                         left += 1
 
                     # we dec the count here so that next round right need to match one more s[left], 
                     d[s[left]] += 1
 
                     if result == "" or len(result) > right - left:
                         result = s[left: right+1]
 
         return result

class Solution:
     def minWindow(self, s, t):
         """
         :type s: str
         :type t: str
         :rtype: str
         """
         table = {}
         for c in t:
             table[c] = table[c] + 1 if c in table else 1
         start, end, count = 0, 0, len(t)
         min_start, min_end = None, None
         for i in range(len(s)):
             if s[i] in table:
                 if table[s[i]] > 0:
                     count -= 1
                 table[s[i]] -= 1
             if count == 0:
                 while count == 0:
                     if s[start] in table:
                         if table[s[start]] >= 0:
                             count += 1
                             if min_start == None or i - start < min_end - min_start:
                                 min_start, min_end = start, i
                         table[s[start]] += 1
                     start += 1
         if min_start == None:
             return ""
         else:
             return s[min_start:(min_end + 1)]
         
         '''
         def check(m, n, mode=1):
             table = dict(zip(t, [0]*len(t)))
             for i in range(m, n+1):
                 if s[i] in table:
                     if mode == 1 and table[s[i]] != None:
                         continue
                     else:
                         table[s[i]] = i
 
             a, b = None, None
             for k, v in table.items():
                 if v == None:
                     return 0, 0, False
                 if a == None or a > v:
                     a = v
                 if b == None or b < v:
                     b = v
             return a, b, True
         
         first, last, result = check(0, len(s)-1, 1)
         if not result:
             return False
         a, b = first, last
         for i in range(last+1, len(s)):
             t_a, t_b, result = check(i-(b-a+1), i, 2)
             if result == True:
                 a, b = t_a, t_b
         
         return s[a:(b+1)]
         '''


class Solution:
     def minWindow(self, s, t):
         """
         :type s: str
         :type t: str
         :rtype: str
         """
         # assume t doesn't have duplicated chars
         left=-1
         right = 0
         result = ""
 
         totalMatch = 0
         d = {}
         for c in t:
             d[c] = d.get(c, 0) + 1
 
         for right in range(len(s)):
             c = s[right]
             if c in d:
                 d[c]-=1
                 if d[c] >=0:
                     # good match
                     totalMatch +=1
 
                     if left == -1:
                         left = right
 
                     #find a sustring
                     if totalMatch == len(t):
                         if result == "" or len(result) > right - left:
                             result = s[left: right+1]
 
                         d[s[left]]+=1
                         totalMatch -=1
                         while left<right:
                             left +=1
 
                             if s[left] in d:
                                 if d[s[left]] <0:
                                     # skip over matched letter
                                     d[s[left]] +=1
                                 else:
                                     break
                 
                 elif c == s[left]:
                     d[c]+=1
                     while left < right:
                         left +=1
                         # skip over matched letter
                         if s[left] in d:
                             if d[s[left]] <0:
                                 d[s[left]] +=1
                             else:
                                 break
         return result

class Solution:
     def minWindow(self, s, t):
         """
         :type s: str
         :type t: str
         :rtype: str
         """
         if len(t) > len(s):
             return ""
         dic = {}
         for i in t:
             if i in dic:
                 dic[i] += 1
             else:
                 dic[i] = 1
         counter = len(dic)
         start = end = 0
         minl = len(s)+1
         res = ""
         
         while end < len(s):
             j = s[end]
             if j in dic:
                 dic[j] -= 1
                 if dic[j] == 0:
                     counter -= 1
             end += 1
             while counter == 0:
                 i = s[start]
                 if i in dic:
                     dic[i] += 1
                     if dic[i] > 0:
                         counter += 1
                     if end-start < minl:
                         minl = end-start
                         res = s[start:end]
                 start += 1
         return res