Smallest distinct window

/*
TC O(N)
Logic:we will have a window and a map,map will store count of all chars in window.
we will keep count of distinct character of window using map and when it equals to
distinct characters in map,then we have a substring with all distinct character
.now there is possibility that some characters got repeated
so we will shrink our window from left dise till our window have every distinct character
then we will update our ans.now we will reduce one distinct character from our window
and search for other possibility and do same for new window.

*/
class Solution{
    public:
    int findSubString(string s){
       set<char> st;
       for(auto itr:s){  //find distinct characters in string
           st.insert(itr);
       }
       int target=st.size();  //total distinct characters in string
       int distinct=0;
       unordered_map<char,int> mp;   //it will store the frequency of characters of window
       int l=0,r=0,ans=INT_MAX,cur=0;
       for(int i=0;i<s.size();i++){
           if(mp[s[i]]!=0){
               mp[s[i]]++;
           }
           else{
               cur++;                 //if frequency is 0,means new char in window
               mp[s[i]]++;            //so it will contribute in distinct characters
               if(cur==target){
                   while(l<s.size() && mp[s[l]]>1){mp[s[l]]--;l++;}  //shrink the window from left
                   ans=min(ans,(i-l+1));
                   mp[s[l]]--;
                   cur--;
                   l++;
               }
           }
       }
       return ans;
    }
};