# 565. 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:

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}

# Solution

Approach 1: just do it -- in O(N) time.

# Code (Python)

Approach 1:

class Solution:
    def arrayNesting(self, nums: List[int]) -> int:
        used = [False] * len(nums)
        max_len = 1
        for current in range(len(nums)):
            current_max_len = 0
            while used[current] == False:
                current_max_len += 1
                used[current] = True
                current = nums[current]
            max_len = max(max_len, current_max_len)
        return max_len

Approach 2:

# Code (C++)

Approach 1:

Approach 2: