How do you solve Linked List Components on LeetCode?

Linked List Components (LeetCode #817) 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: Using a set allows for O(1) lookups, which is crucial for performance.

What is the brute force solution for Linked List Components?

The brute force approach for Linked List Components is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach checks each element in the linked list and counts how many times we find consecutive elements from the nums array. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Linked List Components?

Linked List Components uses the following concepts: Array, Hash Table, Linked List. The recommended patterns to study are: Hash Map, Array.

What are the interview tips for Linked List Components?

Always consider edge cases, such as an empty linked list or nums array. Explain your thought process clearly, especially when transitioning from brute force to optimal solutions. Practice dry runs on paper to solidify your understanding of the algorithm.

What are common mistakes when solving Linked List Components?

Not using a set for quick lookups, leading to unnecessary nested loops. Failing to handle the end of the linked list properly.

#817

Linked List Components

Medium

ArrayHash TableLinked ListHash 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 solution uses a set to quickly check if a node's value is in nums and counts components based on consecutive nodes in the linked list.

⚙️

Algorithm

5 steps

1Step 1: Create a set from the nums array for O(1) lookups.
2Step 2: Initialize a count variable to zero.
3Step 3: Traverse the linked list, checking if each node's value is in the set.
4Step 4: If a value is found, increment the count and skip all consecutive nodes that are also in the set.
5Step 5: Return the count after traversing the entire linked list.

solution.py18 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def numComponents(head, nums):
8    nums_set = set(nums)
9    count = 0
10    current = head
11    while current:
12        if current.val in nums_set:
13            count += 1
14            while current and current.val in nums_set:
15                current = current.next
16        else:
17            current = current.next
18    return count

ℹ

Complexity note: The time complexity is O(n) because we traverse the linked list once, and the space complexity is O(n) due to the set storing the nums values.

1Using a set allows for O(1) lookups, which is crucial for performance.
2Identifying consecutive components is key to solving the problem efficiently.

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