Array Nesting

Problem Statement

Given an integer array nums which has a length of n and is a permutation of the numbers in the range [0, n - 1], you are required to devise sets from this array. These sets, denoted as s[k], begin with an element at index k in nums (nums[k]) and are constructed by continuously indexing into nums using the last value accessed. The construction process of s[k] continues until a value repeats within the set, at which standard the process stops. The challenge here is not merely to build these sets but to determine the maximum length attainable from any of these s[k] sets across all possible 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 <= 105
• 0 <= nums[i] < nums.length
• All the values of nums are unique.

Approach and Intuition

1. Since every nums[k] references another index inside the same array (because nums is a permutation of [0, n - 1]), it forms a cycle.
2. Every number in the array starts a new potential cycle, and given that all entries are unique, there are no cross-links — each number exclusively belongs to a single cycle.
3. Our primary goal is to find the longest such cycle because the length of this cycle will give us the size of our largest set s[k].

Steps to Achieve the Solution:

1. Utilize a depth-first search (DFS) or similar method to traverse each cycle starting from each index that hasn't been visited.
2. Maintain a visited array or set to ensure you don’t recompute for indices already determined to be part of a cycle.
3. Each time you process a new index k, simulate following the chain (nums[k], nums[nums[k]], ...) until you encounter a repeat or reach an already visited index.
4. Record the length of this cycle and compare it with the previously found maximum.
5. The end result after processing all elements will be the maximum length of any set s[k].

Justification from Examples:

• Example 2 Consideration: When there are no cyclic links e.g., nums = [0, 1, 2], every position k refers to itself so the largest set will have a length of 1.
• Example 1 Efficiency: Given an array like [5,4,0,3,1,6,2], the maximum length of a cycle derived from the descriptions is 4 (starting index 0 and following the chain through 5 to 6 to 2 and finally to 0). Multiple cycles may exist but the largest is what we require.

This method leverages the natural cycle properties induced by permutations and ensures that each number is processed efficiently. By avoiding re-examination of numbers and terminating on revisits, it maintains robust performance even for large arrays within the given constraints.

Solutions

public class Solution {
    public int maxArrayNestLength(int[] elements) {
        int maximumLength = 0;
        for (int index = 0; index < elements.length; index++) {
            if (elements[index] != Integer.MAX_VALUE) {
                int element = elements[index], length = 0;
                while (elements[element] != Integer.MAX_VALUE) {
                    int temp = element;
                    element = elements[element];
                    length++;
                    elements[temp] = Integer.MAX_VALUE;
                }
                maximumLength = Math.max(maximumLength, length);
            }
        }
        return maximumLength;
    }
}