What is the brute force solution for Find the Minimum and Maximum Number of Nodes Between Critical Points?

The brute force approach for Find the Minimum and Maximum Number of Nodes Between Critical Points is Brute Force, with O(n²) time complexity and O(1) space complexity. This approach checks every possible pair of critical points to find the minimum and maximum distances. It's straightforward but inefficient, especially for large lists. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Find the Minimum and Maximum Number of Nodes Between Critical Points?

Find the Minimum and Maximum Number of Nodes Between Critical Points uses the following concepts: Linked List. The recommended patterns to study are: Two Pointers, Sliding Window, Array.

What are the interview tips for Find the Minimum and Maximum Number of Nodes Between Critical Points?

Always clarify the problem statement and edge cases before diving into coding. Explain your thought process while coding to showcase your understanding. Test your code with multiple test cases, especially edge cases.

What are common mistakes when solving Find the Minimum and Maximum Number of Nodes Between Critical Points?

Not checking if there are at least two critical points before calculating distances. Confusing the indices of critical points with their values.

#2058

Find the Minimum and Maximum Number of Nodes Between Critical Points

Medium

Linked ListTwo PointersSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

This approach efficiently finds critical points in a single pass and calculates distances in another pass. It reduces unnecessary comparisons and leverages the structure of the linked list.

⚙️

Algorithm

3 steps

1Step 1: Traverse the linked list to identify all critical points and store their indices.
2Step 2: Calculate minDistance by checking the difference between consecutive critical points.
3Step 3: The maxDistance is simply the difference between the first and last critical points.

solution.py26 lines

1# Full working Python code
2class ListNode:
3    def __init__(self, val=0, next=None):
4        self.val = val
5        self.next = next
6
7def nodesBetweenCriticalPoints(head):
8    critical_points = []
9    current = head
10    index = 0
11    while current and current.next and current.next.next:
12        if (current.val < current.next.val > current.next.next.val) or (current.val > current.next.val < current.next.next.val):
13            critical_points.append(index + 1)
14        current = current.next
15        index += 1
16
17    if len(critical_points) < 2:
18        return [-1, -1]
19
20    min_distance = float('inf')
21    max_distance = critical_points[-1] - critical_points[0]
22
23    for i in range(len(critical_points) - 1):
24        min_distance = min(min_distance, critical_points[i + 1] - critical_points[i])
25
26    return [min_distance, max_distance]

ℹ

Complexity note: The time complexity is O(n) because we traverse the linked list once to find critical points and then calculate distances in a single pass. The space complexity is O(n) due to storing the indices of critical points.

1Critical points can only be identified if there are at least three nodes in the list.
2The maximum distance is always between the first and last critical point.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.