algoadvance

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 nums.
• The next element in s[k] is nums[nums[k]], and then nums[nums[nums[k]]], and so on.
• This process stops when a set is formed, meaning the next value to pick already exists in the set.

Return the size of the largest set s[k] formed.

Clarifying Questions

Before solving the problem, let’s clarify a few things:

1. Is the nums array guaranteed to be a permutation of [0, n-1]?
   • Yes, according to the problem statement.
2. Is it a requirement to check for cycles, or can we assume they exist as per the permutation nature?
   • We should assume cycles exist inherently in a permutation, as every permutation of [0, n-1] forms a valid cycle.
3. Can we mutate the original array?
   • Yes, unless stated otherwise, mutating the original array is permissible.

Strategy

To solve this, we’ll employ the following strategy:

1. Iterate through the elements of the array and use each element as a starting point for forming sets.
2. Track visited elements to avoid redundant work.
3. For each unvisited element, traverse through the sequence until we encounter a previously visited element or return to a starting element (thus completing the cycle).
4. Record the size of each set and update the maximum size encountered.

Code

def arrayNesting(nums):
    visited = [False] * len(nums)  # To keep track of visited elements
    max_size = 0  # To keep track of the maximum set size

    for i in range(len(nums)):
        if not visited[i]:
            start = nums[i]
            count = 0
            while not visited[start]:
                visited[start] = True
                start = nums[start]
                count += 1
            max_size = max(max_size, count)

    return max_size

Time Complexity

• Time Complexity: (O(n))
  • Each element in the array is visited exactly once.
• Space Complexity: (O(n))
  • The visited array takes up linear space relative to the input size.

This approach ensures that we efficiently find the maximum set size while keeping track of visited elements to avoid redundant calculations.

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