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 elementnums[k]
ofindex = k
. - The next element in
s[k]
should benums[nums[k]]
, and thennums[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
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