37_Minimum window substring

/*
Remember while solving:
1)Convey your thoughts clearly to the interviewer
2)Maintain topmost code quality
3)Discuss testcases while solving
4)Evolve from brute->Better->optimal
*/

//TC->O(n), where n is the length of string s,Since each character of string s is traversed at most twice, the algorithm’s time complexity is O(n) + O(m).
//SC->O(n) fro hashmap

//link to the problem:https://leetcode.com/problems/minimum-window-substring/
//DSA sheet by Arsh Goyal: Problem 37

//Count frequncy and then track using two pointers

class Solution {
    public String minWindow(String s, String p) {
        HashMap<Character,Integer> map1=new HashMap<>();
        HashMap<Character,Integer> map2=new HashMap<>();
        int p_size=p.length();
        int s_size=s.length();
        int count=0,index=0,sign=1;
        String ans="",final_ans="";
        
        for(int i=0;i<p_size;i++){
            char ch=p.charAt(i);
            map1.put(ch,map1.getOrDefault(ch,0)+1);
        }
        for(int i=0;i<s_size;i++){
            char ch=s.charAt(i);
            map2.put(ch,map2.getOrDefault(ch,0)+1);
            if(map2.getOrDefault(ch,0)<=map1.getOrDefault(ch,0))
                count++;
            if(count==p_size){
                String temp="";
                temp=s.substring(index,i+1);
                
                int flag=1;
                while(flag==1){
                    ch=s.charAt(index);
                    if(map2.get(ch)>map1.getOrDefault(ch,0)){
                        map2.put(ch,map2.get(ch)-1);
                        index++;
                    }
                    else
                        flag=-1;
                }
                ans=s.substring(index,i+1);
                
                if(sign==1){
                    sign=0;
                final_ans=s;
                }
                
                if(final_ans.length()>ans.length())
                    final_ans=ans;
                else if(final_ans.length()>temp.length())
                    final_ans=temp;
                
                map2.put(ch,map2.get(ch)-1);
                count--;
                index++;
            }
        }
        return final_ans;
    }
}