Problem Description

In this problem, we're dealing with a linked list where we need to find critical points - which are defined as either a local maxima or a local minima. A local maxima is considered as a node that has a value strictly greater than the nodes before and after it. Similarly, a local minima is a node that has a value strictly smaller than the nodes before and after it. However, a node can only be considered a critical point if it isn’t the first or last node, meaning it has both a previous and a next node.

Your task is to calculate the minimum and maximum distance between any two critical points in the linked list. The distance is calculated based on the positions of the nodes, not their values. If there are less than two critical points, the answer will be [-1, -1].

Intuition

To solve this problem, we need to traverse through the linked list to find the critical points and calculate the minimum and maximum distances between them.

1. We initiate a traversal starting with the second node in the list since the first node can't be a critical point by the problem's definition.

2. As we traverse, we compare the value of the current node with the values of its previous and next nodes to determine if it's a local maxima or minima.

3. When we identify a critical point, we calculate the distance from the last found critical point. If it’s the first critical point found, we simply mark it as first and last; otherwise, we update the minimum distance every time we find a new critical point, and we set the maximum distance as the difference between the first critical point found and the current one, since this is likely to be the maximum distance as we traverse the list.

4. To keep track of the index of nodes, we use a counter that increments with each traversal step.

5. We utilize a pair of variables [minDistance, maxDistance] initialized to [inf, -inf] to keep track of the distances. These will be updated accordingly as we identify critical points.

6. At the end of traversal, we check whether we have found at least two critical points. If we have only found one or none, we return [-1, -1]. Otherwise, we return the distances found.

By following this approach, we ensure that we only pass through the list once (O(n) time complexity), and use O(1) auxiliary space, ensuring an efficient solution.

Learn more about Linked List patterns.

Solution Approach

The Solution Approach to finding the critical points in a linked list involves a single pass algorithm and judicious use of pointers and variables to keep track of the critical points and their distances.

1. Algorithm: We start by initializing our pointers, prev and curr, to the first and second nodes of the list respectively, because a critical point requires both a previous and next node; therefore, the first node cannot be one.

2. Tracking First and Last Critical Point: We declare variables first and last to remember the positions of the first and last critical points found. They are initially set to None.

3. Initializing the Answer: We prepare an array ans with two elements [inf, -inf] representing the minimum and maximum distances respectively. inf is a placeholder indicating that no minimum distance has been found yet, and -inf indicating that no maximum distance has been found yet.

4. Traversing the List: We iterate through the linked list with a while loop that continues as long as the current node has a next node.

5. Identifying Critical Points:

   • In each iteration, we check if curr.val (the value of the current node) is a local maximum or minimum compared to prev.val and curr.next.val.
   • This is done by checking if curr.val is strictly greater than both prev.val and curr.next.val for a local maximum or strictly smaller for a local minimum.

6. Updating Distances: If a critical point is found:

   • If it’s the first critical point, we initialize both first and last to the current index i.
   • If it's not the first, we:
     • Calculate the distance from the last critical point by subtracting last from the current index i and update ans[0] with the minimum distance found so far.
     • Update ans[1] with the distance between the first critical point and the current one to maintain the maximum distance.
     • Then last is updated to the current index as the last seen critical point.

7. Incrementing Index and Updating Pointers: After checking for a critical point, we increment our index i and move our pointers ahead to the next elements (prev and curr) to continue our traversal.

8. Returning the Result: Once we finish traversing the list:

   • If first and last are the same, it means that either no critical points were found or there was only one critical point. In that case, we return [-1, -1].
   • Otherwise, we return the distances stored in ans, which contain the minimum and maximum distances between critical points.

By applying this step-by-step approach, we efficiently calculate the required distances with straight forward logic and optimal time and space complexity.

Example Walkthrough

Consider the following linked list as an example:

List: 1 -> 3 -> 2 -> 5 -> 3 -> 4

Following our solution approach:

1. We initialize prev to node with value 1 and curr to node with value 3, with an index i starting at 1.

2. Our variables first and last are initialized to None. The array ans is set to [inf, -inf].

3. As we traverse the list, we reach the second node with value 3, it isn’t a critical point because it isn't smaller or larger than both neighbours.

4. Move to the third node with value 2, this is a local minima since 2 is less than both 3 and 5. This is the first critical point, so we set both first and last to the current index i = 2. Since there’s no previous critical point to compare with, we proceed without updating ans.

5. The next node, with value 5, is a local maxima as it is greater than 2 and 3. We find the distance to the last critical point (distance = 3 - 2 = 1), and update the ans array with the minimum distance thus far, ans[0] = min(inf, 1) = 1. We then set last to 3 since 5 is now the last critical point we've seen.

6. Moving forward, node 3 is not a critical point because it isn't a local minima or maxima.

7. Finally, the node with value 4 is a local maxima. We update the minimum distance if it's smaller than the current ans[0] value (it's not in this case: 1 < 2 is false), and then, we update the ans[1] with the maximum distance found which is from the first critical point to the current one (distance = 5 - 2 = 3). We also update last to the current index i = 5.

8. The traversal ends, and since first != last, we have found at least two critical points.

9. We conclude the minimum distance between critical points is 1 and the maximum is 3, hence we return [1, 3].

Here's the example summarized in the required markdown template:

Let's consider a linked list with the following sequence of nodes:

1 -> 3 -> 2 -> 5 -> 3 -> 4

We'll walk through the solution approach step by step.

- We start at the second node (`3`) since the first node can’t be a critical point.
- The second node (`3`) isn’t a local maxima or minima, so we move to the next node.
- At the third node (`2`), we identify our first critical point, it's a local minima.
- Then we come across the fourth node (`5`) which is a local maxima. We find that the minimum distance to the previous critical point is `1`.
- The fifth node (`3`) isn't a critical point, so we continue.
- Lastly, at the sixth node (`4`), we've got another local maxima. We now know the maximum distance is `3`, between the third node and this one.

After walking through the entire linked list, we determine the minimum distance between critical points is `1` and the maximum distance is `3`, which gives us our final answer: `[1, 3]`.

Solution Implementation

Time and Space Complexity

The time complexity of the given code is O(n), where n is the number of nodes in the linked list. The algorithm iterates through the list once with a single pass, checking each node to determine if it's a critical point—either a local minimum or a local maximum. The checks for minimum and maximum at each step are constant-time operations.

The space complexity of the code is O(1). The space used by the algorithm does not scale with the input size as it only maintains a constant number of pointers and variables (prev, curr, first, last, i, and ans) regardless of the length of the list.

Learn more about how to find time and space complexity quickly using problem constraints.