Leetcode 76: Minimum Window Substring

Posted 2020-10-14 Keep walking

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.

 

 1 public class Solution {
 2     public string MinWindow(string s, string t) {
 3         if (s == "" || t == "") return "";
 4         
 5         var dict = new Dictionary<char, int>();
 6         
 7         for (int i = 0; i < t.Length; i++)
 8         {
 9             if (dict.ContainsKey(t[i]))
10             {
11                 dict[t[i]]++;
12             }
13             else
14             {
15                 dict[t[i]] = 1;
16             }
17         }
18         
19         int start = 0, j = 0, count = t.Length;
20         var result = "";
21         
22         while (j < s.Length)
23         {
24             if (dict.ContainsKey(s[j]))
25             {
26                 if (dict[s[j]] > 0)
27                 {
28                     count--;
29                 }
30                 
31                 dict[s[j]]--;
32             }
33             
34             while (count == 0 && start <= j)
35             {
36                 if (result == "" || result.Length > j - start + 1)
37                 {
38                     result = s.Substring(start, j - start + 1);
39                 }
40                 
41                 if (dict.ContainsKey(s[start]))
42                 {
43                     dict[s[start]]++;
44                     
45                     if (dict[s[start]] > 0)
46                     {
47                         count++;
48                     }
49                 }
50                 
51                 start++;
52             }
53             
54             j++;
55         }
56         
57         return result;
58     }
59 }