Leetcode #565: Array Nesting
In this guide, we solve Leetcode #565 Array Nesting in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
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]]], ...
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Depth-First Search, Array
Intuition
We need to explore a structure deeply before backing up, which suits DFS.
DFS keeps local context on the call stack and is easy to implement recursively.
Approach
Define a recursive DFS that carries the necessary state.
Combine child results as the recursion unwinds.
Steps:
- Define a recursive DFS with state.
- Visit children and combine results.
- Return the final aggregation.
Example
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}
Python Solution
class Solution:
def arrayNesting(self, nums: List[int]) -> int:
n = len(nums)
vis = [False] * n
res = 0
for i in range(n):
if vis[i]:
continue
cur, m = nums[i], 1
vis[cur] = True
while nums[cur] != nums[i]:
cur = nums[cur]
m += 1
vis[cur] = True
res = max(res, m)
return res
Complexity
The time complexity is O(V+E). The space complexity is O(V).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.