## DEV Community

Abhishek Chaudhary

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# 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):
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
``````