Untitled

class Solution {
public:
    string minWindow(string s, string t) {
        if(t=="") return "";
        int n=s.size();
        unordered_map<char,int>mt;
        unordered_map<char,int>window;
        int need=t.size(),have=0,res=INT_MAX;
        for(char c:t) mt[c]++;
        int l=0,r=0;
        pair<int,int>ind={-1,-1};

        // mt contains all the counts of characters needed, need has the no of characters needed
        while(r<n){
            char ch=s[r];
            window[ch]++;//map window has count of characters in current window

            if(mt.find(ch)!=mt.end() && window[ch]<=mt[ch]) have++;//current character is needed and our current window has less count than needed then we increment have , we dotn increment have for duplicates

            while(have==need){ //moving the left pointer// improving out answer
                if(r-l+1 <res){
                    res=r-l+1;
                    ind={l,r-l+1};
                }
                
                char left=s[l];
                window[left]--;
                if(mt.find(left)!=mt.end() && window[left] < mt[left]) have--;
                l++;
            }
            r++;
        }
        if(ind.first==-1) return "";
        return s.substr(ind.first,ind.second);
    }
};