class Solution {
    public String minWindow(String s, String t) {
        if(s.length()<t.length())
            return "";

        int[] sChars = new int[256];
        int[] tChars = new int[256];
        int count = 0;
        int start = 0;
        int minLength = Integer.MAX_VALUE;
        int start_index = -1;

        for(char c : t.toCharArray()){
            tChars[c]++;
        }

        for(int i=0;i<s.length();i++){
            sChars[s.charAt(i)]++;
            //Check if this char is present in pattern
            // We check for less than condition only .. lets say s=aaabcd and t=aabc
            // in this case we do not increase the count for 3rd a since that is not required in the pattern
            if(tChars[s.charAt(i)]!=0 && sChars[s.charAt(i)]<=tChars[s.charAt(i)])
                count++;

            if(count==t.length()){
                while(sChars[s.charAt(start)]>tChars[s.charAt(start)] || tChars[s.charAt(start)]==0){
                    //If we are moving the start ahead, we are no longer counting the char at that position
                    //Hence we reduce the count for futher iterations
                    if(sChars[s.charAt(start)]>tChars[s.charAt(start)])
                        sChars[s.charAt(start)]--;

                    start++;
                }
                if(i-start+1<minLength){
                    minLength = i-start+1;
                    start_index = start;
                }
            }
        }
        if(start_index==-1)
            return "";
        return s.substring(start_index,start_index+minLength);

    }
}