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:
[2, 105].1 <= Node.val <= 105In this approach, we traverse the linked list in a single pass, collecting the positions of critical points (local minima and maxima) along the way. We maintain two variables, `firstCritical` and `lastCritical`, to track the position of the first and the last critical points encountered. Additionally, we calculate the minimum distance between consecutive critical points using a running minimum distance variable.
Finally, the maximum distance is the distance between the first and last critical points if at least two critical points are found. Otherwise, return [-1, -1] if fewer than two critical points exist.
We iterate over the linked list using a while loop, checking if the current node is a local maxima or minima by comparing it with its previous and next nodes. We track the indices of the first and last critical points discovered. Whenever another critical point is found, we update the minimum distance using the index difference from the last critical point.
If fewer than two critical points are found, we return [-1, -1]; otherwise, the result is an array of [minDistance, maxDistance].
C++
Java
C
C#
JavaScript
The time complexity of this solution is O(n), where n is the number of nodes in the linked list, because we traverse the list only once. The space complexity is O(1), since we only use a static amount of additional space.
This approach resolves the problem using two full passes over the linked list for clearer data separation. In the first traversal, we gather all the indices of critical points into a list. During the second traversal over the list of critical points, we calculate the minimum and maximum distances between critical point pairs.
This method enhances clarity by separately collecting and then processing data, although it might sacrifice some execution time compared to a single-pass solution.
This Python version uses two loops over a linked list. The first loop extracts indices of critical points, and the second loop computes the desired statistics from them, aiding in understandability and ensuring comprehensive error handling.
The time complexity is O(n) as it traverses the list twice, but the proportional dependence on n remains. Space complexity is O(k), where k is the number of critical points stored.
| Approach | Complexity |
|---|---|
| Single Pass and Collect Critical Points | The time complexity of this solution is O(n), where n is the number of nodes in the linked list, because we traverse the list only once. The space complexity is O(1), since we only use a static amount of additional space. |
| Two-pass Approach for Clarity | The time complexity is O(n) as it traverses the list twice, but the proportional dependence on n remains. Space complexity is O(k), where k is the number of critical points stored. |
Find the Minimum and Maximum Number of Nodes Between Critical Points - Leetcode 2058 - Python • NeetCodeIO • 6,458 views views
Watch 9 more video solutions →Practice Find the Minimum and Maximum Number of Nodes Between Critical Points with our built-in code editor and test cases.
Practice on FleetCodePractice this problem
Open in Editor