Twitter OA prepare: Visit element of the array

A zero-indexed array A consisting of N integers is viven. We visit the indexs of the array in the following way. In the first step we visit the index 0; in every subsequent step we move from the visited index K to the index:
M = K + A[K];
provided M is within the array bounds. Otherwise, K is the last index visited. Write a funciton:
int solution(int A[], int N);
that, given an array A, returns the number of indexes that cannot be visited by the described procedure.
For example, for the array:
A[0] = 1
A[1] = 2
A[2] = 3
only index 2 cannot be visited, so the answer is 1.
For the array:
A[0] = 3
A[1] = -5
A[2] = 0
A[3] = -1
A[4] = -3
indexes 1 and 4 cannot be visited, so the answer is 2.
Assume that:
N is an integer within the range [0...200,000];
each element of array A is an integer within the range [-1,000,000...1,000,000]
Complexity:
expected worst-case time complexity is O(N*log(N));
expected worst-case space complexity is O(N*log(N)), beyond input storage (not counting the storage required for input arguments).
Elements of input arrays can be modified.

 1 public int visiting(int[] A, int N) {
 2     if (A==null || A.length==0) return 0;
 3     int cur = 0;
 4     int count = 0;
 5     boolean[] visited = new boolean[N];
 6     while (cur>=0 && cur<A.length && !visited[cur]) {
 7         visited[cur] = true;
 8         cur = cur + A[cur];
 9         count++;
10     }
11     return N-count;
12 }