Minimum lights to activate

int Solution::solve(vector<int> &A, int B) {
    int n = A.size();
    int count = 0;
    for(int i=0; i< n; i++){
        //We will follow greedy approach.
        //We will go the rightmost light which can light up the current location.
        int next = i+B-1;
        int prev = max(0,i-B+1); //We set the minimum to 0 so that when we do A[prev] prev is not a negative number
        int idx = next; // Going to the rightmost position that can light up the current location.
        bool flag = false; // Boolean to identify if it is possible to light up i at all
        while(idx>=prev){ //prev to idx are the locations that can light up i
            if(A[idx]){
                count++;
                flag = true; // It is possible to light up i
                i = idx+B-1; //This light will be able to light up locations upto idx+B-1
                break; //You have lighted up the ith position. So no need of this while loop anymore
            }
            idx--;
        }
        if(!flag)return -1; //It is not possible to light up the entire street whatsoever.
    }
    return count;
}