<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
"""
1. we can do this in a single pass by keeping track of the minimum and maximum
2. we go through the list and keep track of the minimum distance between critical points as well as the first found critical point and last found critical point
3. at the end take the first found critical point and last found critical point and subtract their distances

edge case:
2 nodes
"""
class Solution:
    def nodesBetweenCriticalPoints(self, head: Optional[ListNode]) -> List[int]:
        first_found = 0
        previous_found = 0
        current_idx = 1
        min_distance = float('inf')
        prev_node = head
        current = head.next
        while current.next:
            if (prev_node.val > current.val and current.val < current.next.val) or (prev_node.val < current.val and current.val > current.next.val):
                if previous_found == 0:
                    previous_found = current_idx
                    first_found = current_idx
                else:
                    min_distance = min(min_distance, current_idx - previous_found)
                    previous_found = current_idx
            current_idx += 1
            prev_node = current
            current = current.next
        if min_distance != float('inf'):
            return [min_distance, previous_found - first_found]
        return [-1,-1]
                

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