835. Linked List Components¶

Difficulty: Medium

835. Linked List Components

Medium

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.

Solution¶

/**
 * 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) {
        HashSet<Integer> set = new HashSet<>();
        for (int ele : nums) set.add(ele);
        ListNode temp = head;
        int count = 0, current_element_count = 0;
        while (temp != null) {
            if (set.size() == 0) break;
            int current_ele = temp.val;
            if (set.contains(current_ele)) {
                set.remove(current_ele);
                current_element_count++;
            }
            else if(current_element_count > 0) {
                count++;
                set.remove(current_ele);
                current_element_count = 0;
            }
            temp = temp.next;
        }
        count++;
        return count;
    }
}

Complexity Analysis¶

Time Complexity: O(?)
Space Complexity: O(?)

Approach¶

Detailed explanation of the approach will be added here