565. Array Nesting

package LeetCode_565

/**
 * 565. Array Nesting
 * https://leetcode.com/problems/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].
 * */
class Solution {
    /*
    * solution: HashSet, Time:O(n), Space:O(n)
    * */
    fun arrayNesting(nums: IntArray): Int {
        val set = HashSet<Int>()
        var max = 0
        for (i in nums.indices) {
            /*
            * for example: [5,4,0,3,1,6,2], set not contains 0, then j = 0,
            * */
            if (!set.contains(i)) {
                var j = i
                var currentCount = 0
                while (!set.contains(j)) {
                    set.add(j)
                    /*
                    * j = nums[0]:j = 5, set not contains 5,
                    * j = nums[5]:j = 6, set not contains 2,
                    * j = nums[2]:j = 0, set contains 0, break while to next for loop
                    * */
                    j = nums[j]
                    currentCount++
                }
                //after break while loop, compare the maximum value
                max = Math.max(max, currentCount)
            }
        }
        return max
    }
}