Introduction

In this blog post, we will explore how to solve the LeetCode problem "817" 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 the head of a linked list containing unique integer values and an integer array nums that is a subset of the linked list values. Return the number of connected components in nums where two values are connected if they appear consecutively in the linked list. Example 1: Input: head = [0,1,2,3], nums = [0,1,3] Output: 2 Explanation: 0 and 1 are connected, so [0, 1] and [3] are the two connected components. Example 2: Input: head = [0,1,2,3,4], nums = [0,3,1,4] Output: 2 Explanation: 0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components. Constraints: The number of nodes in the linked list is n. 1 <= n <= 104 0 <= Node.val < n All the values Node.val are unique. 1 <= nums.length <= n 0 <= nums[i] < n All the values of nums are unique.

Explanation

Here's the solution to the problem:

• High-level approach:

  • Create a set from the nums array for efficient lookups.
  • Iterate through the linked list, checking if the current node's value is in the set.
  • Increment the component count when entering a connected component and only once per component.

• Complexity: O(n) runtime, O(m) storage, where n is the length of the linked list and m is the length of nums.

Code

    # Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def numComponents(self, head: ListNode, nums: list[int]) -> int:
        nums_set = set(nums)
        count = 0
        in_component = False

        current = head
        while current:
            if current.val in nums_set:
                if not in_component:
                    count += 1
                    in_component = True
            else:
                in_component = False
            current = current.next

        return count