What is the time complexity of Self Crossing?

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

What data structures are used to solve Self Crossing?

This problem uses Array, Math, Geometry. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Self Crossing?

Insight 1: The problem can be solved by checking for specific crossing patterns based on the distances traveled. We don't need to track the exact coordinates of the path. Insight 2: There are three main cases where a self-crossing can occur: when the current segment intersects the segment three steps back, when it intersects the segment four steps back, or when it intersects the segment five steps back.

Which companies ask LeetCode 335. Self Crossing in interviews?

This problem is frequently asked at Zomato.

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Self Crossing - LeetCode 335 Solution

Self Crossing - Complete Solution Guide

Self Crossing is LeetCode problem 335, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given an array of integers distance . You start at the point (0, 0) on an X-Y plane, and you move distance[0] meters to the north, then distance[1] meters to the west, distance[2] meters to the south, distance[3] meters to the east, and so on. In other words, after each move, your direction changes counter-clockwise. Return true if your path crosses itself or false if it does not . Example 1: Input: distance = [2,1,1,2] Output: true Explanation: The path crosses itself at the point (0, 1

Detailed Explanation

The problem asks us to determine if a path, formed by a sequence of movements in the north, west, south, and east directions, crosses itself. We start at (0, 0) and move according to the 'distance' array. Each element in the array represents the distance of the next move, and the direction of movement changes counter-clockwise after each move. The goal is to return `true` if the path crosses itself at any point, and `false` otherwise. The constraints are that the length of the distance array is between 1 and 10^5, and each distance value is between 1 and 10^5.

Solution Approach

The solution approach iterates through the distance array, starting from the fourth move (index 3), and checks for three different crossing scenarios. Each scenario represents a different way the path can intersect itself. These scenarios are based on relationships between the lengths of the current segment and the segments preceding it. If any of the crossing conditions are met, the function immediately returns `true`. If the loop completes without finding any crossings, the function returns `false`.

Step-by-Step Algorithm

Step 1: Check if the length of the distance array is less than or equal to 3. If it is, return `false` because a self-crossing is impossible with only a few moves.
Step 2: Iterate through the distance array starting from the fourth element (index 3).
Step 3: Inside the loop, check for the first crossing condition: the current segment crosses the segment three steps before. This happens if `distance[i] >= distance[i-2]` and `distance[i-1] <= distance[i-3]`.
Step 4: If the first condition is not met and the current index `i` is greater than or equal to 4, check for the second crossing condition: the current segment crosses the segment four steps before. This happens if `distance[i-1] == distance[i-3]` and `distance[i] + distance[i-4] >= distance[i-2]`.
Step 5: If the first and second conditions are not met and the current index `i` is greater than or equal to 5, check for the third crossing condition: the current segment crosses the segment five steps before. This happens if `distance[i-2] > distance[i-4]` and `distance[i-3] > distance[i-1]` and `distance[i-1] + distance[i-5] >= distance[i-3]` and `distance[i] + distance[i-4] >= distance[i-2]`.
Step 6: If any of the conditions in steps 3, 4, or 5 are true, return `true`.
Step 7: If the loop completes without returning `true`, return `false`.

Key Insights

Insight 1: The problem can be solved by checking for specific crossing patterns based on the distances traveled. We don't need to track the exact coordinates of the path.
Insight 2: There are three main cases where a self-crossing can occur: when the current segment intersects the segment three steps back, when it intersects the segment four steps back, or when it intersects the segment five steps back.
Insight 3: We can iterate through the `distance` array and check for these crossing conditions based on the previous values.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: Array, Math, Geometry.

Companies

Asked at: Zomato.

Self Crossing | LeetCode 335

Hard

Array Math Geometry

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

Solution Code

Available in:

class Solution {
    public boolean isSelfCrossing(int[] distance) {
        int n = distance.length;
        if (n <= 3) {
            return false;
        }

        for (int i = 3; i < n; i++) {
            // Case 1: The path turns inward and crosses the segment from 3 steps before.
            // The 4th segment crosses the 1st segment.
            if (distance[i] >= distance[i-2] && distance[i-1] <= distance[i-3]) {
                return true;
            }
            
            // Case 2: The path forms a rectangle and the 5th segment touches or crosses
            // the 1st segment. This happens if the path returns to the initial axis.
            // The 5th segment crosses the 1st segment.
            if (i >= 4 && distance[i-1] == distance[i-3] && distance[i] + distance[i-4] >= distance[i-2]) {
                return true;
            }

            // Case 3: The spiral was contracting and then starts expanding, causing
            // the 6th segment to cross the 1st.
            if (i >= 5) {
                if (distance[i-2] > distance[i-4] && distance[i-3] > distance[i-1] &&
                   distance[i-1] + distance[i-5] >= distance[i-3] &&
                   distance[i] + distance[i-4] >= distance[i-2]) {
                    return true;
                }
            }
        }
        
        return false;
    }
}

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Problem Info

Description

Problem 335: Self Crossing - Detailed explanation and solution approaches.

Topics

Array Math Geometry

Companies

Zomato

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:335

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

7 steps

1
Step 1: Check if the length of the distance array is less than or equal to 3. If it is, return `false` because a self-crossing is impossible with only a few moves.
2
Step 2: Iterate through the distance array starting from the fourth element (index 3).
3
Step 3: Inside the loop, check for the first crossing condition: the current segment crosses the segment three steps before. This happens if `distance[i] >= distance[i-2]` and `distance[i-1] <= distance[i-3]`.
4
Step 4: If the first condition is not met and the current index `i` is greater than or equal to 4, check for the second crossing condition: the current segment crosses the segment four steps before. This happens if `distance[i-1] == distance[i-3]` and `distance[i] + distance[i-4] >= distance[i-2]`.
5
Step 5: If the first and second conditions are not met and the current index `i` is greater than or equal to 5, check for the third crossing condition: the current segment crosses the segment five steps before. This happens if `distance[i-2] > distance[i-4]` and `distance[i-3] > distance[i-1]` and `distance[i-1] + distance[i-5] >= distance[i-3]` and `distance[i] + distance[i-4] >= distance[i-2]`.
6
Step 6: If any of the conditions in steps 3, 4, or 5 are true, return `true`.
7
Step 7: If the loop completes without returning `true`, return `false`.

💡

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 O(n) because the solution iterates through the distance array once. The space complexity is O(1) because the solution uses a constant amount of extra space, regardless of the size of the input array. It only uses a few integer variables to store indices and distances.

📖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

Coordinate Tracking

Keep track of the coordinates of each point in the path. After each move, check if the new line segment intersects with any of the previous line segments.

Time: O(n^2)Space: O(n)

Trade-offs: This approach is more intuitive but less efficient than the given solution. The time complexity is higher because for each new segment, you need to check for intersections with all previous segments. The space complexity is higher because you need to store the coordinates of all points.

✨

Key Insights

1Insight 1: The problem can be solved by checking for specific crossing patterns based on the distances traveled. We don't need to track the exact coordinates of the path.
2Insight 2: There are three main cases where a self-crossing can occur: when the current segment intersects the segment three steps back, when it intersects the segment four steps back, or when it intersects the segment five steps back.
3Insight 3: We can iterate through the `distance` array and check for these crossing conditions based on the previous values.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: `distance` array has length less than or equal to 3. The solution handles this by returning `false` at the beginning because a self-crossing requires at least 4 moves.
Edge case 2: A long spiral that eventually crosses back. The solution correctly identifies crossings because it focuses on local crossing conditions rather than tracking the entire path.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Trying to calculate the exact coordinates of each point. This makes the solution unnecessarily complex and prone to errors. It is better to focus on the crossing conditions based on the distances.

✗Common mistake 2: Not considering all three crossing scenarios. The solution correctly identifies the crossing conditions by checking for three separate cases.

✅

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: Robotics and path planning. This algorithm could be used to detect if a robot's planned path will cause it to collide with itself.

🏦Real-world scenario 2: Computer graphics. Detecting self-intersecting lines and curves is essential in rendering objects and preventing visual artifacts.

💡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

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

These problems test your ability to optimize. Start with a brute force solution, then improve.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

👨‍💻 From Experience

I encountered a variation of this problem in a system design interview. The key was recognizing the underlying pattern.

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: November 3, 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

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 13

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJul 6

I was asked this exact problem at my Google interview. The interviewer wanted me to optimize space complexity.

@InterviewProJun 29

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Self Crossing"?

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

🤖 Need help with this problem?