Solution Explanation

This problem asks to find the number of connected components in a subset (nums) of values from a linked list. Two values are considered connected if they appear consecutively in the linked list.

The solution uses a HashSet (or a set in Python) to efficiently check if a value is present in nums. The algorithm iterates through the linked list. It counts a connected component whenever it encounters a node whose value is in nums and it wasn't already in a connected component.

Algorithm:

1. Create a HashSet: Create a HashSet s containing all the values from the input array nums. This allows for O(1) lookup time to check if a node's value is in nums.

2. Iterate the Linked List: Iterate through the linked list using a while loop.

3. Skip Non-members: Skip nodes whose values are not in s.

4. Count Components: When a node with a value in s is found, increment the component count (ans). Then, continue iterating until a node whose value is not in s is encountered.

Time Complexity: O(N), where N is the number of nodes in the linked list. We iterate through the linked list at most once. HashSet operations (insertion and lookup) take O(1) on average.

Space Complexity: O(M), where M is the length of the nums array. This is the space used to store the HashSet s.

Code Explanation (Python):

class Solution:
    def numComponents(self, head: Optional[ListNode], nums: List[int]) -> int:
        ans = 0  # Initialize the component count
        s = set(nums)  # Create a set from nums for efficient lookups
 
        while head:  # Iterate through the linked list
            while head and head.val not in s:  # Skip nodes not in nums
                head = head.next
            ans += head is not None # Increment count if a node in nums is found
            while head and head.val in s:  # Skip nodes in nums that are part of the current component
                head = head.next
        return ans

The code in other languages follows a very similar structure, using their respective set implementations and syntax. The core logic remains the same.