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 emtpy string "".
If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.
The main idea is to maintain two pointers which are called begin and end, then use two HashMap to record current status, one record what char need to find, another used to record what has found.
if the two HashMaps show current substring of S has contain all of the chars in T, then we try to shrink our start and end pointer
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| public class Solution { | |
| public String minWindow(String S, String T) { | |
| if (S==null||T==null){ | |
| return null; | |
| } | |
| if(S.length()==0 && T.length()==0){ | |
| return ""; | |
| } | |
| if (S.length()<T.length()){ | |
| return""; | |
| } | |
| HashMap<Character, Integer>needFind=new HashMap<Character, Integer>(); | |
| HashMap<Character, Integer>alreadyFind=new HashMap<Character, Integer>(); | |
| for(int i=0; i<T.length(); i++){ | |
| alreadyFind.put(T.charAt(i), 0); | |
| if (needFind.containsKey(T.charAt(i))){ | |
| needFind.put(T.charAt(i), needFind.get(T.charAt(i))+1); | |
| } | |
| else{ | |
| needFind.put(T.charAt(i), 1); | |
| } | |
| } | |
| int minStart=-1; | |
| int minEnd=S.length(); | |
| int start=0; | |
| int len=0; | |
| for (int i=0; i<S.length(); i++){ | |
| if (alreadyFind.containsKey(S.charAt(i))){ | |
| alreadyFind.put(S.charAt(i), alreadyFind.get(S.charAt(i))+1); | |
| if (alreadyFind.get(S.charAt(i))<=needFind.get(S.charAt(i))){ | |
| len++; | |
| } | |
| if (len==T.length()){ | |
| while (!needFind.containsKey(S.charAt(start)) || alreadyFind.get(S.charAt(start))>needFind.get(S.charAt(start))){ | |
| if (needFind.containsKey(S.charAt(start))){ | |
| alreadyFind.put(S.charAt(start), alreadyFind.get(S.charAt(start))-1); | |
| } | |
| start++; | |
| } | |
| if (i-start<minEnd-minStart){ | |
| minStart=start; | |
| minEnd=i; | |
| } | |
| } | |
| } | |
| } | |
| if (minStart==-1){ | |
| return ""; | |
| } | |
| return S.substring(minStart, minEnd+1); | |
| } | |
| } |
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