Minimum Window Substring Java

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.

 Key to Solve: Two pointer Window's range method + HashMap,
    Similar idea with Substring with Concatenation of All Words & Longest Substring Without Repeating Characters
    1. Create a Map(key: Character, Value: Frequency) of T
    2. move right window's point while < S.length
    2. move left window's point under condition of full match were found
    3. optimize the min window string
    Time: O(2*N)=O(N)
    Space: O(L) L is size of T

public class Solution {    public String minWindow(String S, String T) {      String output="";    if(S.length()==0 || T.length()==0) return "";    HashMap<Character,Integer> map=new HashMap<Character, Integer>();    int matchCount=0;    int minLen=S.length()+1;    int left=0,right=0;    //create a Map    for(int i=0;i<T.length();i++){        char c=T.charAt(i);        if(map.containsKey(c)){            map.put(c,map.get(c)+1);        }else {            map.put(c,1);        }    }    while(right<S.length()) {        char c = S.charAt(right);        if (map.containsKey(c)) {            //System.out.println("c: "+c);            map.put(c, map.get(c)-1);            if (map.get(c) >= 0) {                matchCount++;            }            //Full Match were found            while (matchCount == T.length()) {                char cl=S.charAt(left);                if (map.containsKey(cl)) {                    //record the character frequency                    map.put(S.charAt(left), map.get(cl) + 1);                    //optimize the min window String                    if (map.get(cl)>0) {                        if (minLen > right - left + 1) {                            output = S.substring(left, right + 1);                            minLen = right - left + 1;                        }                        matchCount--;                    }                }                left++;            }        }        right++;    //skip un-relevant character    }    return output;    }}