Array Nesting

A zero-indexed array A of length N contains all integers from 0 to N-1. Find and return the longest length of set S, where S[i] = {A[i], A[A[i]], A[A[A[i]]], ... } subjected to the rule below.

Suppose the first element in S starts with the selection of element A[i] of index = i, the next element in S should be A[A[i]], and then A[A[A[i]]]… By that analogy, we stop adding right before a duplicate element occurs in S.

 

Example 1:

Input: A = [5,4,0,3,1,6,2]
Output: 4
Explanation: 
A[0] = 5, A[1] = 4, A[2] = 0, A[3] = 3, A[4] = 1, A[5] = 6, A[6] = 2.

One of the longest S[K]:
S[0] = {A[0], A[5], A[6], A[2]} = {5, 6, 2, 0}

 

Note:

  1. N is an integer within the range [1, 20,000].
  2. The elements of A are all distinct.
  3. Each element of A is an integer within the range [0, N-1].

Solution:

class Solution {    
    public int arrayNesting(int[] nums) {
        int n = nums.length;
        int max = 0;
        boolean[] visited = new boolean[n];
        for (int i = 0; i < n; i ++) {
            if (!visited[i]) {
                max = Math.max(max, dfs(nums, i, visited));
            } 
        }
        return max;
    }
    
    private int dfs(int[] nums, int i, boolean[] visited) {
        if (visited[i]) return 0;
        visited[i] = true;
        return 1 + dfs(nums, nums[i], visited);
    }
}