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Array Nesting

MediumUpdated: Aug 2, 2025
Practice on:
LeetCode

Problem

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].

Examples

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 <= 10^5
  • 0 <= nums[i] < nums.length
  • All the values of nums are unique.

Solution

Method 1 – Cycle Detection with Visited Marking

Intuition

The problem is about finding the largest cycle in a permutation array. Since each number is unique and in the range [0, n-1], following the chain from any index will eventually loop back, forming a cycle. By marking visited elements, we avoid redundant work and ensure each cycle is counted only once.

Approach

  1. Initialize a boolean array visited of size n to track visited indices.
  2. For each index i in nums:
  • If i is not visited:
    • Start traversing from i, following the chain nums[i] -> nums[nums[i]] -> ....
    • For each step, mark the current index as visited and increment a counter.
    • Stop when you reach an already visited index (cycle detected).
    • Update the answer if the current cycle length is greater.
  1. Return the maximum cycle length found.

Code

C++
class Solution {
public:
   int arrayNesting(vector<int>& nums) {
      int n = nums.size(), ans = 0;
      vector<bool> vis(n, false);
      for (int i = 0; i < n; ++i) {
        if (!vis[i]) {
           int cnt = 0, j = i;
           while (!vis[j]) {
              vis[j] = true;
              j = nums[j];
              ++cnt;
           }
           ans = max(ans, cnt);
        }
      }
      return ans;
   }
};
Java
class Solution {
   public int arrayNesting(int[] nums) {
      int n = nums.length, ans = 0;
      boolean[] vis = new boolean[n];
      for (int i = 0; i < n; ++i) {
        if (!vis[i]) {
           int cnt = 0, j = i;
           while (!vis[j]) {
              vis[j] = true;
              j = nums[j];
              cnt++;
           }
           ans = Math.max(ans, cnt);
        }
      }
      return ans;
   }
}
Python
class Solution:
   def arrayNesting(self, nums: list[int]) -> int:
      n: int = len(nums)
      vis: list[bool] = [False] * n
      ans: int = 0
      for i in range(n):
        if not vis[i]:
           cnt: int = 0
           j: int = i
           while not vis[j]:
              vis[j] = True
              j = nums[j]
              cnt += 1
           ans = max(ans, cnt)
      return ans

Complexity

  • ⏰ Time complexity: O(n) — Each index is visited at most once.
  • 🧺 Space complexity: O(n) — For the visited array.

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