Find count of good stones (Dynamic Programming) (graph cycles) HARD

https://practice.geeksforgeeks.org/problems/e2d156755ca4e0a9b9abf5680191d4b06e52b1a8/1

Geek is in a geekland which have a river and some stones in it. Initially geek can step on any stone. Each stone has a number on it representing the value of exact step geek can move. If the number is +ve then geeks will move right by arr[i] and if the number is -ve then geeks move left by arr[i]. Bad Stones are defined as the stones in which if geeks steps, will reach a never ending loop whereas good stones are the stones which are safe from never ending loops. Return the number of good stones in river.

Example 1:

Input: [2, 3, -1, 2, -2, 4, 1]
Output: 3
Explanation: Index 3, 5 and 6 are safe only. As index 1, 4, 2 forms a cycle and from index 0 you can go to index 2 which is part of cycle.

Example 2:

Input: [1, 0, -3, 0, -5, 0]
Output: 2
Explanation: Index 2 and 4 are safe only. As index 0, 1, 3, 5 form cycle.

Expected Time Complexity : O(N), N is the number of stones
Expected Auxiliary Space : O(N), N is the number of stones

Constraints:
   1 <= n < 10^5 (where n is the length of the array)
  -1000 <= arr[i] < 1000
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class Solution{
public:
    bool isBad(vector<int> &arr, int curr, vector<string> &states){
        if(curr>=arr.size() || curr<0 || states[curr]=="GOOD"){
            return false;  // not bad
        }
        if(states[curr]=="PROCESSING" || states[curr]=="BAD"){
            return true; // bad
        }
        states[curr]="PROCESSING"; // NODE UNDER PROCESS
        bool answer=isBad(arr,curr+arr[curr],states);
        if(answer==true){
            states[curr]="BAD";
        }
        else{
            states[curr]="GOOD";
        }
        return answer;
    }

    int goodStones(int n,vector<int> &arr){
        vector<string> states(n,"NULL");
        for(int i=0;i<n;i++){
            if(states[i]=="NULL"){
                isBad(arr,i,states);
            }
        }
        int res=0;
        for(auto c: states){
            if(c=="GOOD"){
                res++;
            }
        }
        return res;
    }
};