Array Nesting

You are given an integer array nums of length n where nums is a permutation of the numbers in the range [0, n - 1].

You should build a set s[k] = {nums[k], nums[nums[k]], nums[nums[nums[k]]], ... } subjected to the following rule:

• The first element in s[k] starts with the selection of the element nums[k] of index = k.
• The next element in s[k] should be nums[nums[k]], and then nums[nums[nums[k]]], and so on.
• We stop adding right before a duplicate element occurs in s[k].

Return the longest length of a set s[k].

Example 1:

Input: nums = [5,4,0,3,1,6,2]
Output: 4
Explanation:
nums[0] = 5, nums[1] = 4, nums[2] = 0, nums[3] = 3, nums[4] = 1, nums[5] = 6, nums[6] = 2.
One of the longest sets s[k]:
s[0] = {nums[0], nums[5], nums[6], nums[2]} = {5, 6, 2, 0}

Example 2:

Input: nums = [0,1,2]
Output: 1

Constraints:

• 1 <= nums.length <= 105
• 0 <= nums[i] < nums.length
• All the values of nums are unique.

SOLUTION:

class Solution:
    def DFS(self, nums, i, visited):
        visited.add(i)
        if nums[i] not in visited:
            return self.DFS(nums, nums[i], visited)
        else:
            return (i, visited)

    def arrayNesting(self, nums: List[int]) -> int:
        n = len(nums)
        mlen = 0
        visited = set()
        for i in range(n):
            if i not in visited:
                start, l = self.DFS(nums, i, set())
                end, l = self.DFS(nums, start, set())
                visited.update(l)
                mlen = max(mlen, len(l))
        return mlen