Problem 2058 Find the Minimum and Maximum Number of Nodes Between Critical Points
Table of Contents
Problem Statement
Link - Problem 2058
Question
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
graph LR
A((3)) --> B((1))
Input: head = [3,1]
Output: [-1,-1]
Explanation: There are no critical points in [3,1].
Example 2
graph LR
A((5)) --> B((3))
B --> C((1))
C --> D((2))
D --> E((5))
E --> F((1))
F --> G((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
graph LR
A((1)) --> B((3))
B --> C((2))
C --> D((2))
D --> E((3))
E --> F((2))
F --> G((2))
G --> H((2))
H --> I((7))
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, 10^5]`.
- `1 <= Node.val <= 10^5`
Solution
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode() : val(0), next(nullptr) {}
* ListNode(int x) : val(x), next(nullptr) {}
* ListNode(int x, ListNode *next) : val(x), next(next) {}
* };
*/
class Solution {
public:
vector<int> nodesBetweenCriticalPoints(ListNode* head) {
std::ios::sync_with_stdio(false);
cin.tie(nullptr);
cout.tie(nullptr);
ListNode* prev = head;
ListNode* curr = head->next;
if(head->next == nullptr)
return {-1,-1};
vector<int>positions;
int ind = 1;
while(curr->next!=nullptr)
{
if((prev->val<curr->val && curr->val>curr->next->val)||(prev->val>curr->val && curr->val<curr->next->val))
{
positions.push_back(ind);
}
ind++;
prev = curr;
curr =curr->next;
}
if(positions.size()<2)
return {-1,-1};
int mini = INT_MAX;
for(int i = 1;i<positions.size();i++)
mini = min(mini,positions[i]-positions[i-1]);
return {mini,positions.back() - positions.front()};
}
};
Complexity Analysis
Time Complexity: O(n)Space Complexity: O(1) excluding the space for the output list.
Explanation
1. Intuition
- We need to find the critical points in the linked list and then find the minimum and maximum distance between them.
- That means first we have to find the indexes of the critical points.
- Then we can find the minimum and maximum distance between them.
- We can find the critical points by comparing the current node with the previous and the next node.
- Create an ascending sequence of the indexes of the critical points.
- Maximum will be the difference between the last and the first index.
- Minimum will be the minimum difference between the indexes.
- Edge cases are when there are no critical points or only one critical point.
- When there is less than 3 nodes in the linked list, there will be no critical points.
2. Implementation
- Delcare two pointers `prev` and `curr` to traverse the linked list.
- `prev` will point to the `head` and `curr` will point to the next node.
- If there are less than 2 nodes in the linked list, return `[-1,-1]`.
- Declare a vector `positions` to store the indexes of the critical points.
- Declare an integer `ind` to store the index of the current node, initialize it to 1.
- While `curr->next` is not `nullptr`
1. If the current node is a critical point, add the index to the `positions` vector.
2. Increment the index `ind`.
3. Move the `prev` and `curr` pointers to the next node.
- If there are less than 2 critical points, return `[-1,-1]`.
- Initialize an integer `mini` to `INT_MAX`.
- Iterate over the `positions` vector from the second element to the last element.
1. Find the minimum difference between the indexes of the critical points.
- Return the minimum and maximum distance between the critical points.