<p>A <strong>critical point</strong> in a linked list is defined as <strong>either</strong> a <strong>local maxima</strong> or a <strong>local minima</strong>.</p>

<p>A node is a <strong>local maxima</strong> if the current node has a value <strong>strictly greater</strong> than the previous node and the next node.</p>

<p>A node is a <strong>local minima</strong> if the current node has a value <strong>strictly smaller</strong> than the previous node and the next node.</p>

<p>Note that a node can only be a local maxima/minima if there exists <strong>both</strong> a previous node and a next node.</p>

<p>Given a linked list <code>head</code>, return <em>an array of length 2 containing </em><code>[minDistance, maxDistance]</code><em> where </em><code>minDistance</code><em> is the <strong>minimum distance</strong> between <strong>any&nbsp;two distinct</strong> critical points and </em><code>maxDistance</code><em> is the <strong>maximum distance</strong> between <strong>any&nbsp;two distinct</strong> critical points. If there are <strong>fewer</strong> than two critical points, return </em><code>[-1, -1]</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/13/a1.png" style="width: 148px; height: 55px;" />
<pre>
<strong>Input:</strong> head = [3,1]
<strong>Output:</strong> [-1,-1]
<strong>Explanation:</strong> There are no critical points in [3,1].
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/13/a2.png" style="width: 624px; height: 46px;" />
<pre>
<strong>Input:</strong> head = [5,3,1,2,5,1,2]
<strong>Output:</strong> [1,3]
<strong>Explanation:</strong> There are three critical points:
- [5,3,<strong><u>1</u></strong>,2,5,1,2]: The third node is a local minima because 1 is less than 3 and 2.
- [5,3,1,2,<u><strong>5</strong></u>,1,2]: The fifth node is a local maxima because 5 is greater than 2 and 1.
- [5,3,1,2,5,<u><strong>1</strong></u>,2]: The sixth node is a local minima because 1 is less than 5 and 2.
The minimum distance is between the fifth and the sixth node. minDistance = 6 - 5 = 1.
The maximum distance is between the third and the sixth node. maxDistance = 6 - 3 = 3.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/10/14/a5.png" style="width: 624px; height: 39px;" />
<pre>
<strong>Input:</strong> head = [1,3,2,2,3,2,2,2,7]
<strong>Output:</strong> [3,3]
<strong>Explanation:</strong> There are two critical points:
- [1,<u><strong>3</strong></u>,2,2,3,2,2,2,7]: The second node is a local maxima because 3 is greater than 1 and 2.
- [1,3,2,2,<u><strong>3</strong></u>,2,2,2,7]: The fifth node is a local maxima because 3 is greater than 2 and 2.
Both the minimum and maximum distances are between the second and the fifth node.
Thus, minDistance and maxDistance is 5 - 2 = 3.
Note that the last node is not considered a local maxima because it does not have a next node.
</pre>

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

<ul>
	<li>The number of nodes in the list is in the range <code>[2, 10<sup>5</sup>]</code>.</li>
	<li><code>1 &lt;= Node.val &lt;= 10<sup>5</sup></code></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 nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
        # Initialize variables to store the indices of critical points
        critical_points = []
        
        # Initialize variables for tracking the previous node and current index
        prev_node = head
        curr_node = head.next
        curr_index = 1
        
        # Traverse the linked list to find critical points
        while curr_node.next:
            next_node = curr_node.next
            
            # Check if the current node is a local maxima or minima
            if (prev_node.val < curr_node.val and curr_node.val > next_node.val) or \
               (prev_node.val > curr_node.val and curr_node.val < next_node.val):
                critical_points.append(curr_index)
            
            # Move to the next node
            prev_node = curr_node
            curr_node = next_node
            curr_index += 1
        
        # If there are fewer than 2 critical points, return [-1, -1]
        if len(critical_points) < 2:
            return [-1, -1]
        
        # Calculate the minimum and maximum distances between critical points
        min_distance = float('inf')
        max_distance = critical_points[-1] - critical_points[0]
        
        for i in range(1, len(critical_points)):
            min_distance = min(min_distance, critical_points[i] - critical_points[i - 1])
        
        return [min_distance, max_distance]
379