Introduction

In this blog post, we will explore how to solve the LeetCode problem "565" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

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]]], ... } 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.

Explanation

Here's the solution to find the longest length of a set s[k] given the specified conditions:

• High-Level Approach:

  • Iterate through the nums array. For each index k, trace the sequence starting from nums[k] until a duplicate is found.
  • To avoid redundant calculations, use a visited array to keep track of the elements already part of a cycle. If an element is visited, skip the sequence starting from that element.
  • Maintain a max_length variable to store the maximum length of the cycle found so far.

• Complexity:

  • Runtime Complexity: O(n) - Each element is visited at most once.
  • Storage Complexity: O(n) - For the visited array.

Code

    def array_nesting(nums):
    n = len(nums)
    visited = [False] * n
    max_length = 0

    for i in range(n):
        if not visited[i]:
            length = 0
            current = i
            while not visited[current]:
                visited[current] = True
                current = nums[current]
                length += 1
            max_length = max(max_length, length)

    return max_length