[Leetcode] Two pointer-- 76. Minimum Window Substring

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).

For example,
S = "ADOBECODEBANC"
T = "ABC"

Minimum window is "BANC".

Note:
If there is no such window in S that covers all characters in T, return the empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

 

Solution:

windows problem: use two pointer, fix left and move right pointer, traverse source s and find the current windows containing t,
then move the left point to next point to update the current window length to get the minimum window

 1 targetHash = {}         #store the target character counter     
 2         
 3         for c in t:
 4             if c not in targetHash:
 5                 targetHash[c] = 1
 6             else:
 7                 targetHash[c] += 1
 8         
 9         srcHash = {}
10         
11         count = 0         #decide when T matched in S
12         left = 0
13         right = 0
14         minWindLen = 2**32
15         minStart = 0
16         
17         #print ("target hash: ", targetHash)
18         while (right < len(s)):
19             
20             if s[right] in targetHash and targetHash[s[right]] > 0:
21                 if s[right] not in srcHash:
22                     srcHash[s[right]] = 1
23                 else:
24                      srcHash[s[right]] += 1
25                 if (srcHash[s[right]] <= targetHash[s[right]]):
26                     count += 1
27             
28             if count == len(t):
29                     
30                 
31                 while (s[left] not in targetHash  or (s[left] in srcHash and s[left] in targetHash and srcHash[s[left]] > targetHash[s[left]])):
32                     if s[left] in targetHash:
33                         srcHash[s[left]] -= 1
34                     left += 1
35                     
36                 if minWindLen > (right-left+1):
37                     minWindLen = right-left+1
38                     minStart = left
39                                 
40             
41             right += 1
42                 
43         return s[minStart: minStart+minWindLen] if minWindLen <= len(s) else ""

 

posted @ 2018-01-09 09:13  安新  阅读(119)  评论(0)  编辑  收藏  举报