What is the time complexity of Count Collisions on a Road?

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

What data structures are used to solve Count Collisions on a Road?

This problem uses String, Stack, Simulation. Understanding these concepts is essential for solving this Medium problem efficiently.

What are the key insights for solving Count Collisions on a Road?

Insight 1: Cars moving left at the beginning and cars moving right at the end will never collide, as they are moving away from the main traffic flow. Insight 2: Focus on the cars in the 'middle' section of the road. Once you identify the range of cars that *can* collide, the collisions are simply the number of non-stationary cars within that range.

Which companies ask LeetCode 2211. Count Collisions on a Road in interviews?

This problem is frequently asked at Arcesium.

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Count Collisions on a Road - LeetCode 2211 Solution

Count Collisions on a Road - Complete Solution Guide

Count Collisions on a Road is LeetCode problem 2211, 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

There are n cars on an infinitely long road. The cars are numbered from 0 to n - 1 from left to right and each car is present at a unique point. You are given a 0-indexed string directions of length n . directions[i] can be either 'L' , 'R' , or 'S' denoting whether the i th car is moving towards the left , towards the right , or staying at its current point respectively. Each moving car has the same speed . The number of collisions can be calculated as follows: When two cars moving in opposite

Detailed Explanation

The problem presents a scenario of cars moving on an infinitely long road. Each car is either moving left ('L'), right ('R'), or stationary ('S'). The task is to calculate the total number of collisions based on the following rules: 1) Opposite directions colliding result in 2 collisions. 2) A moving car colliding with a stationary car results in 1 collision. After a collision, the involved cars become stationary. The input is a string 'directions' representing the movement of each car, and the output is the total number of collisions.

Solution Approach

The provided solution efficiently calculates the number of collisions by first identifying the boundaries of the potentially colliding cars. It ignores the leftmost cars moving left and the rightmost cars moving right as these cars will not collide with anything. Then, it iterates through the remaining cars and counts the number of cars that are not stationary ('S'). This count directly corresponds to the number of collisions because each such car will inevitably collide with either another moving car (resulting in 2 collisions which is simplified to 1 as the net effect is only 1 collision added), or a stationary car (resulting in 1 collision).

Step-by-Step Algorithm

Step 1: Initialize 'left' to 0. Iterate from the beginning of the 'directions' string, incrementing 'left' until a non-'L' character is encountered or the end of the string is reached.
Step 2: Initialize 'right' to n-1, where n is the length of 'directions'. Iterate from the end of the string, decrementing 'right' until a non-'R' character is encountered or 'left' exceeds the index.
Step 3: Initialize 'collisions' to 0.
Step 4: Iterate from 'left' to 'right' (inclusive). For each character in 'directions' within this range, if the character is not 'S', increment 'collisions'.
Step 5: Return the final value of 'collisions'.

Key Insights

Insight 1: Cars moving left at the beginning and cars moving right at the end will never collide, as they are moving away from the main traffic flow.
Insight 2: Focus on the cars in the 'middle' section of the road. Once you identify the range of cars that *can* collide, the collisions are simply the number of non-stationary cars within that range.
Insight 3: After a collision, cars become stationary. This simplifies the problem because we don't need to track individual car states after each collision. We just count the total number of cars that are not stationary within the collision-prone range.

Complexity Analysis

Time Complexity: O(n)

Space Complexity: O(1)

Topics

This problem involves: String, Stack, Simulation.

Companies

Asked at: Arcesium.

Count Collisions on a Road | LeetCode 2211

Medium

String Stack Simulation

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

Solution Code

Available in:

class Solution {
    public int countCollisions(String directions) {
        int n = directions.length();
        
        int left = 0;
        while (left < n && directions.charAt(left) == 'L') {
            left++;
        }
        
        int right = n - 1;
        while (right >= left && directions.charAt(right) == 'R') {
            right--;
        }
        
        int collisions = 0;
        for (int i = left; i <= right; i++) {
            if (directions.charAt(i) != 'S') {
                collisions++;
            }
        }
        
        return collisions;
    }
}

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Problem Info

Description

Problem 2211: Count Collisions on a Road - Detailed explanation and solution approaches.

Topics

String Stack Simulation

Companies

Arcesium

Stats

Difficulty:Medium

Languages:4

Companies:1

Problem #:2211

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

5 steps

1
Step 1: Initialize 'left' to 0. Iterate from the beginning of the 'directions' string, incrementing 'left' until a non-'L' character is encountered or the end of the string is reached.
2
Step 2: Initialize 'right' to n-1, where n is the length of 'directions'. Iterate from the end of the string, decrementing 'right' until a non-'R' character is encountered or 'left' exceeds the index.
3
Step 3: Initialize 'collisions' to 0.
4
Step 4: Iterate from 'left' to 'right' (inclusive). For each character in 'directions' within this range, if the character is not 'S', increment 'collisions'.
5
Step 5: Return the final value of 'collisions'.

💡

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 linear scans of the input string. The space complexity is constant because only a fixed number of variables are used, irrespective of the input size. The algorithm does not use any auxiliary data structures that scale with the input.

📖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

Stack-based approach

An alternative approach could use a stack to simulate the car movements and track collisions. Iterate through the 'directions' string. If the stack is empty, push the current direction. If the stack contains 'R' and the current direction is 'L', then pop 'R' from the stack, increment the collision count by 2, and push 'S' onto the stack. Handle other cases (R-S, S-L, S-R, L-S) similarly. This method directly simulates the collision process.

Time: O(n)Space: O(n) in the worst case where the stack stores almost all the chars in input string, for example all 'R's

Trade-offs: While conceptually intuitive, the stack-based approach has a higher space complexity (O(n)) compared to the O(1) space complexity of the provided solution. The stack-based approach might be more suitable if more detailed collision information (e.g., collision order) is needed, but for simply counting collisions, the provided solution is more efficient in terms of space.

✨

Key Insights

1Insight 1: Cars moving left at the beginning and cars moving right at the end will never collide, as they are moving away from the main traffic flow.
2Insight 2: Focus on the cars in the 'middle' section of the road. Once you identify the range of cars that *can* collide, the collisions are simply the number of non-stationary cars within that range.
3Insight 3: After a collision, cars become stationary. This simplifies the problem because we don't need to track individual car states after each collision. We just count the total number of cars that are not stationary within the collision-prone range.

📚

Key Terms Explained

For Students

Stack

LIFO structure - Last In, First Out. Like a stack of plates!

String

A sequence of characters. Like a word or sentence in programming!

Edge Cases

Edge case 1: The string contains only 'L' or 'R' characters. In this case, 'left' will reach the end of the string or 'right' will reach the beginning, resulting in an empty range to iterate over in the final loop, and correctly returning 0 collisions.
Edge case 2: The string contains only 'S' characters. In this case, 'left' and 'right' won't be updated, so the loop iterates through the whole array and skips every character, returning 0. Also correctly handled.
Edge case 3: Empty string. While not explicitly disallowed by the constraints, the code gracefully handles it by resulting in left > right and no iteration, resulting in correct 0 collision.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Incorrectly handling edge cases where all cars are moving in the same direction (all 'L' or all 'R'). The initial while loops are crucial to skip these cars and avoid unnecessary calculations.

✗Common mistake 2: Overcomplicating the collision logic by trying to track individual car states after each collision. The problem statement simplifies things by stating that collided cars become stationary, meaning only the count of non-stationary cars within the collision range matters.

✗Common mistake 3: Not considering the edge case where all cars are stationary.

✅

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: Traffic simulation. This algorithm can be used as a simplified model within a larger traffic simulation to estimate the number of collisions in a given road segment based on the direction of traffic flow.

🏦Real-world scenario 2: Robotics. In a warehouse with multiple robots moving along a defined path, this type of collision detection logic (with appropriate modifications for varying robot speeds and positions) can be used to predict potential collisions and adjust robot paths to prevent them.

💡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

String problems often involve sliding windows or two pointers. Consider if the character set is limited (lowercase letters = only 26 possibilities).

🎯 Interview Tip

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

⚠️ Watch Out

Mutating strings directly in languages where strings are immutable. Use StringBuilder or arrays.

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: July 22, 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

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

@AlgoMasterJul 14

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

@InterviewProJul 7

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

@CodeNinjaJun 30

❓

Frequently Asked Questions

What is the sliding window technique for string problems?

Sliding window maintains a "window" of characters with two pointers. Expand the right pointer to include characters, shrink from left when window violates constraints. Useful for substring problems, anagrams, and longest/shortest substring with properties.

What is a monotonic stack and when should I use it?

A monotonic stack maintains elements in sorted order (increasing or decreasing). Use it for "next greater element," "largest rectangle," or "stock span" problems. It achieves O(n) time by avoiding redundant comparisons through maintaining order invariants.

How much time should I spend on a medium problem in interviews?

In a 45-minute interview, aim for: 5-10 minutes understanding and clarifying, 5-10 minutes discussing approach, 20-25 minutes coding, and 5-10 minutes testing. If stuck after 15 minutes of thinking, verbalize your thought process and ask for hints.

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 "Count Collisions on a Road"?

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

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