LeetCode:Minimum Window Substring

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 emtpy string "".

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

import java.util.HashMap;
import java.util.Map;


public class Solution {
    public String minWindow(String S, String T) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(S==null || T==null) return "";
        Map<Character,Integer> total = new HashMap<Character,Integer>();
        for(int i=0;i<T.length();i++){
            char c = T.charAt(i);
            if(total.containsKey(c))
                total.put(c,total.get(c)+1);
            else
                total.put(c,1);
        } 
        
        Map<Character,Integer> covered = new HashMap<Character,Integer>();
        int start=0,
            end=-1,
            cur1=0,
            cur2=0,
            count=0;

        while(cur1<S.length() && cur2<S.length()){
            while(cur1<S.length() && !total.containsKey(S.charAt(cur1))) 
                cur1++;
                
            if(cur1<S.length()){
                if(cur2<cur1) cur2=cur1;
                
                if(count<T.length()){
                    while(cur2<S.length() && !total.containsKey(S.charAt(cur2))) 
                        cur2++;
                    if(cur2<S.length()){
                        char c = S.charAt(cur2);   
                        if(!covered.containsKey(c) ){
                            covered.put(c,1);
                            count++;
                        }else{                        
                            if(covered.get(c)<total.get(c)) 
                                count++;
                            covered.put(c,covered.get(c)+1);
                        } 
                        if(count<T.length()) cur2++;
                    }
                }else{
                    if(start>end || end-start>cur2-cur1){
                        end=cur2;
                        start=cur1;
                    }
                    char c = S.charAt(cur1);
                    covered.put(c,covered.get(c)-1);
                    cur1++;
                    if(covered.get(c)<total.get(c)){
                         count--;
                         cur2++;
                    }
                }
            }
        }
        return  S.substring(start,end+1);
    }
}