What is the time complexity of Find the Minimum and Maximum Number of Nodes Between Critical Points?

The optimal solution has a time complexity of O(n).

What data structures are used to solve Find the Minimum and Maximum Number of Nodes Between Critical Points?

This problem uses Linked List. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Find the Minimum and Maximum Number of Nodes Between Critical Points?

Insight 1: Iterate through the linked list, keeping track of the previous and next nodes to identify local maxima and minima. Insight 2: Maintain the index of the first critical point found and the index of the previously found critical point to efficiently compute distances.

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Find the Minimum and Maximum Number of Nodes Between Critical Points - LeetCode 2058 Solution

Q: Which companies ask LeetCode 2058. Find the Minimum and Maximum Number of Nodes Between Critical Points in interviews?

This problem is frequently asked at Info Edge, josh technology.

Find the Minimum and Maximum Number of Nodes Between Critical Points - Complete Solution Guide

Find the Minimum and Maximum Number of Nodes Between Critical Points is LeetCode problem 2058, a Medium level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

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 [minDis

Detailed Explanation

The problem asks us to find the minimum and maximum distance between any two distinct critical points in a singly linked list. A critical point is defined as a local maxima or a local minima. A local maxima is a node whose value is strictly greater than its previous and next nodes. A local minima is a node whose value is strictly smaller than its previous and next nodes. We need to return an array containing the minimum and maximum distances. If there are fewer than two critical points, we should return [-1, -1]. The input is the head of the linked list, and the output is an array of length 2.

Solution Approach

The solution iterates through the linked list, keeping track of the previous node, current node, and the index of the current node. For each node, it checks if it is a local maxima or minima by comparing its value to the values of its previous and next nodes. If a critical point is found, its index is recorded. The minimum and maximum distances between critical points are calculated by comparing the current critical point's index with the index of the previous critical point. The algorithm updates the minimum distance as necessary. Finally, the maximum distance between the first and last critical point is calculated.

Step-by-Step Algorithm

Step 1: Initialize `min_distance` to infinity, `first_critical_idx` and `prev_critical_idx` to -1, and the index counter to 2 (since the first node cannot be a critical point).
Step 2: Iterate through the linked list from the second node until the second to last node (because the last node also can't be a critical point).
Step 3: In each iteration, check if the current node is a local maxima or minima.
Step 4: If the current node is a critical point, update `first_critical_idx` if it's the first critical point found. Otherwise, calculate the distance between the current critical point and the previous one, and update `min_distance` with the minimum of current `min_distance` and the calculated distance.
Step 5: Update `prev_critical_idx` to the index of the current critical point.
Step 6: After the loop, if `min_distance` is still infinity, return [-1, -1] because there are less than two critical points.
Step 7: Calculate `max_distance` as the difference between the last critical point and the first critical point.
Step 8: Return the array containing `min_distance` and `max_distance`.

Key Insights

Insight 1: Iterate through the linked list, keeping track of the previous and next nodes to identify local maxima and minima.
Insight 2: Maintain the index of the first critical point found and the index of the previously found critical point to efficiently compute distances.
Insight 3: Handle the edge case where there are fewer than two critical points by returning [-1, -1].

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Linked List.

Companies

Asked at: Info Edge, josh technology.

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

Medium

Linked List

Time: O(n)Space: O(1)

Solution Code

Available in:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public int[] nodesBetweenCriticalPoints(ListNode head) {
        if (head == null || head.next == null || head.next.next == null) {
            return new int[]{-1, -1};
        }

        int min_distance = Integer.MAX_VALUE;
        
        int first_critical_idx = -1;
        int prev_critical_idx = -1;
        
        ListNode prev_node = head;
        ListNode curr_node = head.next;
        int index = 2;

        while (curr_node.next != null) {
            ListNode next_node = curr_node.next;
            
            boolean is_maxima = curr_node.val > prev_node.val && curr_node.val > next_node.val;
            boolean is_minima = curr_node.val < prev_node.val && curr_node.val < next_node.val;
            
            if (is_maxima || is_minima) {
                if (first_critical_idx == -1) {
                    first_critical_idx = index;
                    prev_critical_idx = index;
                } else {
                    min_distance = Math.min(min_distance, index - prev_critical_idx);
                    prev_critical_idx = index;
                }
            }
            
            prev_node = curr_node;
            curr_node = next_node;
            index += 1;
            
        }
        
        if (min_distance == Integer.MAX_VALUE) {
            return new int[]{-1, -1};
        }
        
        int max_distance = prev_critical_idx - first_critical_idx;
        
        return new int[]{min_distance, max_distance};
    }
}

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Problem Info

Description

Problem 2058: Find the Minimum and Maximum Number of Nodes Between Critical Points - Detailed explanation and solution approaches.

Topics

Linked List

Companies

Info Edgejosh technology

Stats

Difficulty:Medium

Languages:4

Companies:2

Problem #:2058

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

8 steps

1
Step 1: Initialize `min_distance` to infinity, `first_critical_idx` and `prev_critical_idx` to -1, and the index counter to 2 (since the first node cannot be a critical point).
2
Step 2: Iterate through the linked list from the second node until the second to last node (because the last node also can't be a critical point).
3
Step 3: In each iteration, check if the current node is a local maxima or minima.
4
Step 4: If the current node is a critical point, update `first_critical_idx` if it's the first critical point found. Otherwise, calculate the distance between the current critical point and the previous one, and update `min_distance` with the minimum of current `min_distance` and the calculated distance.
5
Step 5: Update `prev_critical_idx` to the index of the current critical point.
6
Step 6: After the loop, if `min_distance` is still infinity, return [-1, -1] because there are less than two critical points.
7
Step 7: Calculate `max_distance` as the difference between the last critical point and the first critical point.
8
Step 8: Return the array containing `min_distance` and `max_distance`.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(n)

👍 Linear - Good! Visits each element once.

💾

Space Complexity

O(1)

💪 Constant space - Uses same memory regardless of input!

📝

Why?

The time complexity is determined by the single pass through the linked list. The space complexity is constant because the number of variables used does not depend on the size of the input list.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Using an Array to Store Critical Point Indices

Instead of maintaining the first and previous critical point indices, we could store all the indices of critical points in an array. Then, we can iterate through the array to calculate the minimum and maximum distances.

Time: O(n), where n is the number of nodes in the linked list.Space: O(k), where k is the number of critical points. In the worst case, k can be close to n.

Trade-offs: This approach uses more space, especially if there are many critical points. However, it might be slightly easier to understand and implement if you are not comfortable with updating indices on the fly.

✨

Key Insights

1Insight 1: Iterate through the linked list, keeping track of the previous and next nodes to identify local maxima and minima.
2Insight 2: Maintain the index of the first critical point found and the index of the previously found critical point to efficiently compute distances.
3Insight 3: Handle the edge case where there are fewer than two critical points by returning [-1, -1].

📚

Key Terms Explained

For Students

Linked List

Elements connected via pointers. Like a chain where each link knows the next.

Edge Cases

Edge case 1: Empty linked list or linked list with only one or two nodes. The solution handles this case at the beginning by checking if head, head.next or head.next.next is null and returns [-1, -1].
Edge case 2: Linked list with no critical points. The solution handles this by checking if min_distance remains infinity after iterating through the entire linked list. If it does, it means no two critical points were found, and it returns [-1, -1].
Edge case 3: Linked list with only two critical points. The solution correctly calculates the minimum and maximum distance as the distance between those two points.
Edge case 4: All nodes have the same value, resulting in no critical points. Handled correctly because the `is_maxima` and `is_minima` conditions will never be true.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not handling the edge case where there are fewer than two critical points, resulting in incorrect output.

✗Common mistake 2: Incorrectly identifying local maxima and minima by not considering the strict inequality condition ('>' and '<' instead of '>=' and '<=').

✗Common mistake 3: Not updating the previous critical point index correctly after finding a new critical point, leading to incorrect distance calculations.

✗Common mistake 4: Forgetting to check boundary conditions, like making sure the current node has both a previous and a next node before checking for critical points.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Analyzing stock prices to identify local peaks and troughs, which can be considered critical points. The distances between these points can indicate trends and patterns.

🏦Real-world scenario 2: Signal processing to identify local maxima and minima in a signal waveform. The distances between these points can be used to analyze the frequency and amplitude of the signal.

🎮Real-world scenario 3: Monitoring sensor data (e.g., temperature, pressure) to detect significant changes or anomalies. Critical points can represent moments of rapid increase or decrease, and their separation can provide insights into the system's behavior.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

The dummy node technique prevents edge cases. For cycle detection, Floyd's algorithm is the standard. Always handle empty list and single node cases.

🎯 Interview Tip

Medium problems are the bread and butter of technical interviews. Practice explaining your approach before coding.

⚠️ Watch Out

Losing reference to nodes when modifying pointers. Use temporary variables to save references.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: August 18, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

Try solving this without using extra space first. It's a great exercise for understanding the underlying pattern.

@CodeNinjaJul 14

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 7

❓

Frequently Asked Questions

Why use a dummy node in linked list problems?

A dummy node simplifies edge cases where the head might change (insertion/deletion at head) or when building a new list. Initialize dummy.next = head, work with dummy.next throughout, and return dummy.next at the end to handle all cases uniformly.

What techniques are commonly tested in medium-difficulty problems?

Medium problems often require combining multiple techniques: two-pointer with hash maps, BFS/DFS with state tracking, dynamic programming with optimization, or divide-and-conquer approaches. They test your ability to recognize patterns and choose the right combination of techniques.

How should I practice algorithm problems effectively?

Focus on understanding patterns rather than memorizing solutions. After solving a problem, review optimal solutions and understand the intuition. Group similar problems to recognize patterns. Practice explaining your approach out loud. Review problems after a few days to test retention.

What is the optimal approach for "Find the Minimum and Maximum Number of Nodes Between Critical Points"?

For this medium-level Linked List problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether standard techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?