<p>You are given the <code>head</code> of a linked list containing unique integer values and an integer array <code>nums</code> that is a subset of the linked list values.</p>

<p>Return <em>the number of connected components in </em><code>nums</code><em> where two values are connected if they appear <strong>consecutively</strong> in the linked list</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/22/lc-linkedlistcom1.jpg" style="width: 424px; height: 65px;" />
<pre>
<strong>Input:</strong> head = [0,1,2,3], nums = [0,1,3]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 0 and 1 are connected, so [0, 1] and [3] are the two connected components.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/07/22/lc-linkedlistcom2.jpg" style="width: 544px; height: 65px;" />
<pre>
<strong>Input:</strong> head = [0,1,2,3,4], nums = [0,3,1,4]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 0 and 1 are connected, 3 and 4 are connected, so [0, 1] and [3, 4] are the two connected components.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the linked list is <code>n</code>.</li>
	<li><code>1 &lt;= n &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= Node.val &lt; n</code></li>
	<li>All the values <code>Node.val</code> are <strong>unique</strong>.</li>
	<li><code>1 &lt;= nums.length &lt;= n</code></li>
	<li><code>0 &lt;= nums[i] &lt; n</code></li>
	<li>All the values of <code>nums</code> are <strong>unique</strong>.</li>
</ul>

# 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:
        
        rtn = 0
        prev = False

        while head:

            if head.val in nums:
                nums.remove( head.val )

                if not prev:
                    prev = True
                    rtn += 1

            else:
                prev = False
            
            head = head.next

        return rtn

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