Leetcode 817: Linked List Components

Input: The input consists of two parts: a linked list and an array nums. The linked list contains unique integer values, and nums contains integers that are a subset of the linked list's values.

Example: Input: head = [4,5,6,7,8], nums = [4,5,7]

Constraints:

• 1 <= n <= 10^4

• 0 <= Node.val < n

• All the values in Node.val are unique.

• 1 <= nums.length <= n

• 0 <= nums[i] < n

• All the values of nums are unique.

Output: Return the number of connected components, where two values are connected if they appear consecutively in the linked list.

Example: Output: 2

Constraints:

• The number of connected components should be an integer.

Goal: The goal is to traverse the linked list and count the number of connected components where consecutive numbers in nums appear in the linked list.

Steps:

• Create a set from nums to facilitate fast lookups.

• Traverse the linked list while checking for consecutive values that appear in nums.

• Each time a transition from a number in nums to a non-consecutive value occurs, count a new connected component.

Goal: The linked list will contain unique values, and nums is a subset of those values.

Steps:

• The linked list length can be up to 10,000 nodes.

• The values in the linked list are unique.

Assumptions:

• The input linked list contains unique integer values.

• The nums array contains integers that are a subset of the linked list's values.

• Input: Input: head = [0,1,2,3], nums = [0,1,3]

• Explanation: In this case, the linked list is [0,1,2,3] and nums is [0,1,3]. The first connected component is [0, 1], and the second one is [3], so the output is 2.

• Input: Input: head = [4,5,6,7,8], nums = [4,5,7]

• Explanation: Here, the linked list is [4,5,6,7,8] and nums is [4,5,7]. The connected components are [4, 5] and [7], so the output is 2.

Approach: We can solve this problem by traversing the linked list and checking whether the consecutive nodes in the linked list are part of nums. We will count each group of consecutive nodes that are in nums as a connected component.

Observations:

• We need to check consecutive nodes that belong to nums, so we can leverage a set for quick lookups.

• Whenever a node in nums is followed by a non-nums value, we count that as a new component.

• The problem involves identifying consecutive connected nodes in a linked list that belong to the nums array.

Steps:

• Convert nums into a set for O(1) lookups.

• Traverse the linked list. For each node, check if it's in nums and if the next node is also in nums.

• Each time a node from nums is not followed by another node from nums, increment the count of connected components.

Empty Inputs:

• If the input linked list is empty, return 0 as there are no connected components.

Large Inputs:

• For large linked lists (up to 10,000 nodes), ensure that the solution runs efficiently within time limits.

Special Values:

• Handle cases where nums has only one element or contains all elements from the linked list.

Constraints:

• Ensure the solution efficiently handles up to 10,000 nodes in the linked list.

int numComponents(ListNode* head, vector<int>& nums) {
    unordered_set<int> setG(nums.begin(), nums.end());
    int res =0;
    while(head != NULL) {
        if(setG.count(head->val) && 
        (head->next == NULL || !setG.count(head->next->val)))
        res++;
        head = head->next;
    }
    return res;
}

1 : Function Definition

int numComponents(ListNode* head, vector<int>& nums) {

This line defines the function `numComponents` which takes a linked list `head` and a vector of integers `nums` as inputs.

2 : Data Structure Initialization

    unordered_set<int> setG(nums.begin(), nums.end());

An unordered set `setG` is created from the vector `nums`, which will be used to quickly check if a node's value is part of the set.

3 : Variable Initialization

    int res =0;

The variable `res` is initialized to 0, which will be used to count the number of connected components.

4 : Loop Start

    while(head != NULL) {

The loop starts and will iterate through each node of the linked list as long as `head` is not NULL.

5 : Condition Check

        if(setG.count(head->val) &&

This condition checks if the current node's value is in the set `setG`.

6 : Condition Check

        (head->next == NULL || !setG.count(head->next->val)))

This condition checks if the next node is either NULL (end of the list) or its value is not in the set `setG`.

7 : Increment Count

        res++;

If the conditions are met, it indicates the end of a connected component, and `res` is incremented.

8 : Advance to Next Node

        head = head->next;

The `head` pointer is advanced to the next node in the linked list.

9 : Return Result

    return res;

The function returns the value of `res`, which represents the number of connected components.

Best Case: O(n), where n is the length of the linked list.

Average Case: O(n), as each node is processed once.

Worst Case: O(n), since we only traverse the list once and check membership in a set.

Description: The time complexity is linear in terms of the size of the linked list.

Best Case: O(m), where m is the size of nums.

Worst Case: O(m), where m is the size of nums, as we store nums in a set for efficient lookups.

Description: The space complexity is linear in terms of the size of nums due to the set used for fast lookups.

Leetcode 817: Linked List Components

Solution to LeetCode 817: Linked List Components Problem

Explore →