LeetCode 565: Array Nesting Solution in Python – A Step-by-Step Guide

What Is LeetCode 565: Array Nesting?

In LeetCode 565: Array Nesting, you’re given an array nums of length n where each element nums[i] is an integer between 0 and n-1, representing an index to jump to. Starting from any index, you form a sequence by following these jumps (e.g., nums[i] → nums[nums[i]]), which eventually loops back into a cycle. Your task is to return the length of the longest possible sequence (cycle length plus any lead-in). For example, with nums = [5, 4, 0, 3, 1, 6, 2], starting at index 0 gives [0, 5, 6, 2] (length 4). This problem builds on LeetCode 141: Linked List Cycle for cycle detection but applies it to array indices.

Problem Statement

• Input: nums (List[int])—array of integers where 0 <= nums[i] < n.
• Output: Integer—length of the longest nesting sequence.
• Rules: Each element is an index; sequence follows nums[i] → nums[nums[i]]; find max cycle length.

Constraints

• 1 <= nums.length <= 10⁵
• 0 <= nums[i] < nums.length
• Values may repeat, forming cycles.

Examples

• Input: nums = [5,4,0,3,1,6,2]
• Output: 4
• Sequence from 0: [0, 5, 6, 2] cycles back to 0, length = 4.
• Input: nums = [0,1,2]
• Output: 1
• Each index points to itself: [0], [1], [2], max length = 1.
• Input: nums = [1,0]
• Output: 2
• Sequence: [0, 1] cycles back, length = 2.

Understanding the Problem: Finding the Longest Cycle

To solve LeetCode 565: Array Nesting in Python, we need to find the longest sequence formed by following indices in an array, where each element nums[i] points to the next index, eventually forming a cycle. With lengths up to 10⁵, a brute-force approach checking all starting points and sequences could work but might repeat effort (O(n²)). Since each index leads to a cycle, and all starting points in a cycle yield the same length, we can optimize by detecting cycles efficiently and tracking visited indices. We’ll explore:

• Best Solution (Cycle Detection with Visited Tracking): O(n) time, O(n) space—fast and optimal.
• Alternative Solution (Brute-Force Cycle Enumeration): O(n²) time, O(1) space—simple but inefficient.

Let’s dive into the cycle detection solution with a friendly breakdown!

Best Solution: Cycle Detection with Visited Tracking

Why Cycle Detection Wins

The cycle detection with visited tracking is the best for LeetCode 565 because it finds the longest sequence in O(n) time and O(n) space by traversing the array once, using a visited set to mark seen indices and compute cycle lengths only for unvisited starting points, avoiding redundant work. It’s like following a treasure map, marking your path, and counting the longest loop you discover—all in one efficient trip!

How It Works

Think of this as a cycle-hunting adventurer:

• Step 1: Initialize Visited Set:
• Use a set to track seen indices.
• Step 2: Traverse from Each Start:
• For each unvisited index, follow jumps until a cycle (revisit an index).
• Step 3: Count Cycle Length:
• Track length of sequence; once cycle detected, it’s the max for that loop.
• Step 4: Update Max Length:
• Compare with global max, update if larger.
• Step 5: Return Result:
• Longest sequence length.
• Why It Works:
• Each index belongs to one cycle; visiting it once is enough.
• Visited set ensures O(n) total steps.

It’s like a cycle-length explorer!

Step-by-Step Example

Example: nums = [5, 4, 0, 3, 1, 6, 2]

• Init: visited = set(), max_length = 0.
• Step 1: Start at 0 (unvisited):
• 0 → 5 → 6 → 2 → 0 (cycle).
• Length = 4, visited = {0, 5, 6, 2}, max_length = 4.
• Step 2: Start at 1 (unvisited):
• 1 → 4 → 1 (cycle).
• Length = 2, visited = {0, 5, 6, 2, 1, 4}, max_length = 4.
• Step 3: Start at 3 (unvisited):
• 3 → 3 (cycle).
• Length = 1, visited = {0, 5, 6, 2, 1, 4, 3}, max_length = 4.
• Step 4: All visited, stop.
• Result: 4.

Example: nums = [0, 1, 2]

• Step 1: Start at 0:
• 0 → 0, length = 1, max_length = 1.
• Step 2: Start at 1:
• 1 → 1, length = 1, max_length = 1.
• Step 3: Start at 2:
• 2 → 2, length = 1, max_length = 1.
• Result: 1.

Code with Detailed Line-by-Line Explanation

Here’s the Python code, explained for beginners:

Copy codeclass Solution:
    def arrayNesting(self, nums: List[int]) -> int:
        # Step 1: Initialize visited set and max length
        visited = set()
        max_length = 0
        n = len(nums)

        # Step 2: Iterate each starting index
        for start in range(n):
            if start not in visited:
                # Step 3: Follow jumps and count length
                length = 0
                curr = start
                while curr not in visited:
                    visited.add(curr)
                    curr = nums[curr]
                    length += 1
                # Step 4: Update max length
                max_length = max(max_length, length)

        # Step 5: Return maximum length
        return max_length

• Lines 4-6: Set up visited set and max tracker.
• Lines 9-17: For each unvisited start:
• Follow jumps, count length until cycle.
• Mark indices visited.
• Line 19: Update max_length with longest cycle.
• Line 22: Return result.
• Time Complexity: O(n)—each index visited once.
• Space Complexity: O(n)—visited set.

It’s like a cycle-length champion!

Alternative Solution: Brute-Force Cycle Enumeration

Why an Alternative Approach?

The brute-force cycle enumeration checks every starting index, follows the sequence to count its length, and tracks the maximum, running in O(n²) time and O(1) space (excluding minimal temp storage). It’s simple but redundant, making it a good baseline for small arrays or understanding.

How It Works

Picture this as a cycle-counting wanderer:

• Step 1: Iterate all starting indices.
• Step 2: Follow jumps, count until cycle or end.
• Step 3: Track max length.
• Step 4: Return max.

It’s like a cycle-length roamer!

Step-by-Step Example

Example: nums = [1, 0, 2]

• Step 1: Start at 0:
• 0 → 1 → 0, length = 2, max_length = 2.
• Step 2: Start at 1:
• 1 → 0 → 1, length = 2, max_length = 2.
• Step 3: Start at 2:
• 2 → 2, length = 1, max_length = 2.
• Result: 2.

Code for Brute-Force Approach

Copy codeclass Solution:
    def arrayNesting(self, nums: List[int]) -> int:
        n = len(nums)
        max_length = 0

        for start in range(n):
            length = 0
            curr = start
            seen = set()
            while curr not in seen:
                seen.add(curr)
                curr = nums[curr]
                length += 1
                if length > n:  # Avoid infinite loop (though constrained)
                    break
            max_length = max(max_length, length)

        return max_length

• Lines 6-15: For each start, count sequence length, track max.
• Line 17: Return max length.
• Time Complexity: O(n²)—each start may traverse full array.
• Space Complexity: O(1)—minimal extra space.

It’s a brute-force cycle counter!

Comparing the Two Solutions

• Cycle Detection (Best):
• Pros: O(n), O(n), fast.
• Cons: Extra space.
• Brute-Force (Alternative):
• Pros: O(n²), O(1), simple.
• Cons: Slower.

Cycle detection wins for efficiency!

Additional Examples and Edge Cases

• [0]: 1.
• [0, 2, 1]: 2.
• [3, 2, 1, 0]: 4.

Cycle detection handles them all!

Complexity Recap

• Cycle Detection: Time O(n), Space O(n).
• Brute-Force: Time O(n²), Space O(1).

Cycle detection’s the speed champ!

Key Takeaways

• Cycle Detection: Smart tracking—learn at Python Basics!
• Brute-Force: Full roam.
• Arrays: Nesting is fun.
• Python: Detect or brute, your pick!

Final Thoughts: Find That Nest!

LeetCode 565: Array Nesting in Python is a clever array challenge. Cycle detection with visited tracking is your fast track, while brute-force offers a thorough dive. Want more? Try LeetCode 141: Linked List Cycle or LeetCode 287: Find the Duplicate Number. Ready to nest? Head to Solve LeetCode 565 on LeetCode and find that longest sequence today!