LeetCode #817 — MEDIUM

Linked List Components

Move from brute-force thinking to an efficient approach using array strategy.

The Problem

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.

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Array · Hash Map · Linked List

Example 1

[0,1,2,3]
[0,1,3]

Example 2

[0,1,2,3,4]
[0,3,1,4]

Core Insight

What unlocks the optimal approach

No official hints in dataset. Start from constraints and look for a monotonic or reusable state.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #817: Linked List Components
/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int numComponents(ListNode head, int[] nums) {
        int ans = 0;
        Set<Integer> s = new HashSet<>();
        for (int v : nums) {
            s.add(v);
        }
        while (head != null) {
            while (head != null && !s.contains(head.val)) {
                head = head.next;
            }
            ans += head != null ? 1 : 0;
            while (head != null && s.contains(head.val)) {
                head = head.next;
            }
        }
        return ans;
    }
}

// Accepted solution for LeetCode #817: Linked List Components
/**
 * Definition for singly-linked list.
 * type ListNode struct {
 *     Val int
 *     Next *ListNode
 * }
 */
func numComponents(head *ListNode, nums []int) int {
	s := map[int]bool{}
	for _, v := range nums {
		s[v] = true
	}
	ans := 0
	for head != nil {
		for head != nil && !s[head.Val] {
			head = head.Next
		}
		if head != nil {
			ans++
		}
		for head != nil && s[head.Val] {
			head = head.Next
		}
	}
	return ans
}

# Accepted solution for LeetCode #817: Linked List Components
# 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

// Accepted solution for LeetCode #817: Linked List Components
// Definition for singly-linked list.
// #[derive(PartialEq, Eq, Clone, Debug)]
// pub struct ListNode {
//   pub val: i32,
//   pub next: Option<Box<ListNode>>
// }
//
// impl ListNode {
//   #[inline]
//   fn new(val: i32) -> Self {
//     ListNode {
//       next: None,
//       val
//     }
//   }
// }
use std::collections::HashSet;
impl Solution {
    pub fn num_components(head: Option<Box<ListNode>>, nums: Vec<i32>) -> i32 {
        let set = nums.into_iter().collect::<HashSet<i32>>();
        let mut res = 0;
        let mut in_set = false;
        let mut cur = &head;
        while let Some(node) = cur {
            if set.contains(&node.val) {
                if !in_set {
                    in_set = true;
                    res += 1;
                }
            } else {
                in_set = false;
            }
            cur = &node.next;
        }
        res
    }
}

// Accepted solution for LeetCode #817: Linked List Components
/**
 * Definition for singly-linked list.
 * class ListNode {
 *     val: number
 *     next: ListNode | null
 *     constructor(val?: number, next?: ListNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.next = (next===undefined ? null : next)
 *     }
 * }
 */

function numComponents(head: ListNode | null, nums: number[]): number {
    const set = new Set<number>(nums);
    let res = 0;
    let cur = head;
    let inSet = false;
    while (cur != null) {
        if (set.has(cur.val)) {
            if (!inSet) {
                inSet = true;
                res++;
            }
        } else {
            inSet = false;
        }
        cur = cur.next;
    }
    return res;
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n)

Space

O(1)

Approach Breakdown

COPY TO ARRAY

O(n) time

O(n) space

Copy all n nodes into an array (O(n) time and space), then use array indexing for random access. Operations like reversal or middle-finding become trivial with indices, but the O(n) extra space defeats the purpose of using a linked list.

IN-PLACE POINTERS

O(n) time

O(1) space

Most linked list operations traverse the list once (O(n)) and re-wire pointers in-place (O(1) extra space). The brute force often copies nodes to an array to enable random access, costing O(n) space. In-place pointer manipulation eliminates that.

Shortcut: Traverse once + re-wire pointers → O(n) time, O(1) space. Dummy head nodes simplify edge cases.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Off-by-one on range boundaries

Wrong move: Loop endpoints miss first/last candidate.

Usually fails on: Fails on minimal arrays and exact-boundary answers.

Fix: Re-derive loops from inclusive/exclusive ranges before coding.

Mutating counts without cleanup

Wrong move: Zero-count keys stay in map and break distinct/count constraints.

Usually fails on: Window/map size checks are consistently off by one.

Fix: Delete keys when count reaches zero.

Losing head/tail while rewiring

Wrong move: Pointer updates overwrite references before they are saved.

Usually fails on: List becomes disconnected mid-operation.

Fix: Store next pointers first and use a dummy head for safer joins.