Introduction

In this blog post, we will explore how to solve the LeetCode problem "2058" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

A critical point in a linked list is defined as either a local maxima or a local minima. A node is a local maxima if the current node has a value strictly greater than the previous node and the next node. A node is a local minima if the current node has a value strictly smaller than the previous node and the next node. Note that a node can only be a local maxima/minima if there exists both a previous node and a next node. Given a linked list head, return an array of length 2 containing [minDistance, maxDistance] where minDistance is the minimum distance between any two distinct critical points and maxDistance is the maximum distance between any two distinct critical points. If there are fewer than two critical points, return [-1, -1]. Example 1: Input: head = [3,1] Output: [-1,-1] Explanation: There are no critical points in [3,1]. Example 2: Input: head = [5,3,1,2,5,1,2] Output: [1,3] Explanation: There are three critical points: - [5,3,1,2,5,1,2]: The third node is a local minima because 1 is less than 3 and 2. - [5,3,1,2,5,1,2]: The fifth node is a local maxima because 5 is greater than 2 and 1. - [5,3,1,2,5,1,2]: 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. Example 3: Input: head = [1,3,2,2,3,2,2,2,7] Output: [3,3] Explanation: There are two critical points: - [1,3,2,2,3,2,2,2,7]: The second node is a local maxima because 3 is greater than 1 and 2. - [1,3,2,2,3,2,2,2,7]: 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. Constraints: The number of nodes in the list is in the range [2, 105]. 1 <= Node.val <= 105

Explanation

Here's a breakdown of the solution:

• Identify Critical Points: Traverse the linked list, checking each node (excluding the head and tail) to see if it's a local maxima or minima. Store the indices of these critical points.
• Calculate Distances: If there are at least two critical points, iterate through the list of critical points and calculate the minimum and maximum distances between them.

• Handle Edge Cases: If there are fewer than two critical points, return [-1, -1].

• Runtime Complexity: O(N), where N is the number of nodes in the linked list.

• Storage Complexity: O(K), where K is the number of critical points (in the worst case, it can be O(N)).

Code

    class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

def criticalPoint(head):
    critical_points = []
    index = 1
    prev = head
    curr = head.next

    if not curr:
        return [-1, -1]

    next_node = curr.next

    while next_node:
        if curr.val > prev.val and curr.val > next_node.val:
            critical_points.append(index + 1)
        elif curr.val < prev.val and curr.val < next_node.val:
            critical_points.append(index + 1)

        prev = curr
        curr = next_node
        next_node = next_node.next
        index += 1

    if len(critical_points) < 2:
        return [-1, -1]

    min_distance = float('inf')
    max_distance = 0

    for i in range(1, len(critical_points)):
        distance = critical_points[i] - critical_points[i - 1]
        min_distance = min(min_distance, distance)
        max_distance = max(max_distance, critical_points[i] - critical_points[0])

    return [min_distance, max_distance]