Leetcode #817: Linked List Components
In this guide, we solve Leetcode #817 Linked List Components in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Array, Hash Table, Linked List
Intuition
Fast membership checks and value lookups are the heart of this problem, which makes a hash map the natural choice.
By storing what we have already seen (or counts/indexes), we can answer the question in one pass without backtracking.
Approach
Scan the input once, using the map to detect when the condition is satisfied and to update state as you go.
This keeps the solution linear while remaining easy to explain in an interview setting.
Steps:
- Initialize a hash map for seen items or counts.
- Iterate through the input, querying/updating the map.
- Return the first valid result or the final computed value.
Example
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.
Python Solution
# 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: Optional[ListNode], nums: List[int]) -> int:
ans = 0
s = set(nums)
while head:
while head and head.val not in s:
head = head.next
ans += head is not None
while head and head.val in s:
head = head.next
return ans
Complexity
The time complexity is O(n). The space complexity is O(n).
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.