DEV Community

Cover image for 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points
MD ARIFUL HAQUE
MD ARIFUL HAQUE

Posted on • Edited on

1

2058. Find the Minimum and Maximum Number of Nodes Between Critical Points

2058. Find the Minimum and Maximum Number of Nodes Between Critical Points

Medium

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:

a1

  • Input: head = [3,1]
  • Output: [-1,-1]
  • Explanation: There are no critical points in [3,1].

Example 2:

a2

  • 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:

a5

  • 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

Solution:

/**
 * Definition for a singly-linked list.
 * class ListNode {
 *     public $val = 0;
 *     public $next = null;
 *     function __construct($val = 0, $next = null) {
 *         $this->val = $val;
 *         $this->next = $next;
 *     }
 * }
 */
class Solution {

    /**
     * @param ListNode $head
     * @return Integer[]
     */
    function nodesBetweenCriticalPoints($head) {
        $result = [-1, -1];

        // Initialize minimum distance to the maximum possible value
        $minDistance = PHP_INT_MAX;

        // Pointers to track the previous node, current node, and indices
        $previousNode = $head;
        $currentNode = $head->next;
        $currentIndex = 1;
        $previousCriticalIndex = 0;
        $firstCriticalIndex = 0;

        while ($currentNode->next != null) {
            // Check if the current node is a local maxima or minima
            if (($currentNode->val < $previousNode->val &&
                    $currentNode->val < $currentNode->next->val) ||
                ($currentNode->val > $previousNode->val &&
                    $currentNode->val > $currentNode->next->val)) {
                // If this is the first critical point found
                if ($previousCriticalIndex == 0) {
                    $previousCriticalIndex = $currentIndex;
                    $firstCriticalIndex = $currentIndex;
                } else {
                    // Calculate the minimum distance between critical points
                    $minDistance = min($minDistance, $currentIndex - $previousCriticalIndex);
                    $previousCriticalIndex = $currentIndex;
                }
            }

            // Move to the next node and update indices
            $currentIndex++;
            $previousNode = $currentNode;
            $currentNode = $currentNode->next;
        }

        // If at least two critical points were found
        if ($minDistance != PHP_INT_MAX) {
            $maxDistance = $previousCriticalIndex - $firstCriticalIndex;
            $result = [$minDistance, $maxDistance];
        }

        return $result;
    }
}
Enter fullscreen mode Exit fullscreen mode

Contact Links

If you found this series helpful, please consider giving the repository a star on GitHub or sharing the post on your favorite social networks 😍. Your support would mean a lot to me!

If you want more helpful content like this, feel free to follow me:

Image of AssemblyAI tool

Transforming Interviews into Publishable Stories with AssemblyAI

Insightview is a modern web application that streamlines the interview workflow for journalists. By leveraging AssemblyAI's LeMUR and Universal-2 technology, it transforms raw interview recordings into structured, actionable content, dramatically reducing the time from recording to publication.

Key Features:
🎥 Audio/video file upload with real-time preview
🗣️ Advanced transcription with speaker identification
⭐ Automatic highlight extraction of key moments
✍️ AI-powered article draft generation
📤 Export interview's subtitles in VTT format

Read full post

Top comments (0)

AWS Security LIVE!

Tune in for AWS Security LIVE!

Join AWS Security LIVE! for expert insights and actionable tips to protect your organization and keep security teams prepared.

Learn More

👋 Kindness is contagious

Discover a treasure trove of wisdom within this insightful piece, highly respected in the nurturing DEV Community enviroment. Developers, whether novice or expert, are encouraged to participate and add to our shared knowledge basin.

A simple "thank you" can illuminate someone's day. Express your appreciation in the comments section!

On DEV, sharing ideas smoothens our journey and strengthens our community ties. Learn something useful? Offering a quick thanks to the author is deeply appreciated.

Okay