How do you solve Array Nesting on LeetCode?

Array Nesting (LeetCode #565) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Each index can only be part of one unique set due to the permutation property.

What is the brute force solution for Array Nesting?

The brute force approach for Array Nesting is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves exploring each starting index and following the sequence until we encounter a duplicate. This method is straightforward but inefficient as it may revisit elements multiple times. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Array Nesting?

Array Nesting uses the following concepts: Array, Depth-First Search. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Array Nesting?

Always consider edge cases, such as the smallest and largest possible inputs. Explain your thought process clearly; this shows your understanding of the problem. Practice dry-running your code on paper to catch logical errors before coding.

What are common mistakes when solving Array Nesting?

Not using a visited array, leading to repeated calculations. Confusing the set length calculation with the number of unique elements.

#565

Array Nesting

Medium

ArrayDepth-First SearchHash MapArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach leverages the fact that each index can only be part of one set. By marking visited indices, we can efficiently calculate the length of each set without revisiting elements.

⚙️

Algorithm

3 steps

1Step 1: Initialize a visited array to keep track of indices that have been processed.
2Step 2: For each index, if it hasn't been visited, start a new set and follow the sequence until a duplicate is found.
3Step 3: Update the maximum length based on the length of the current set.

solution.py13 lines

1def arrayNesting(nums):
2    visited = [False] * len(nums)
3    max_length = 0
4    for i in range(len(nums)):
5        if not visited[i]:
6            current_length = 0
7            current_index = i
8            while not visited[current_index]:
9                visited[current_index] = True
10                current_index = nums[current_index]
11                current_length += 1
12            max_length = max(max_length, current_length)
13    return max_length

ℹ

Complexity note: The time complexity is O(n) because each index is processed only once, leading to linear time. The space complexity is O(n) due to the visited array that tracks which indices have been processed.

1Each index can only be part of one unique set due to the permutation property.
2Tracking visited indices allows for efficient computation of set lengths.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.