What is the time complexity of Robot Collisions?

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

What data structures are used to solve Robot Collisions?

This problem uses Array, Stack, Sorting, Simulation. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Robot Collisions?

Insight 1: Sorting by position is crucial to process collisions in the correct order as they happen on the number line. Insight 2: A stack is an effective data structure for keeping track of robots moving right ('R') and waiting for potential collisions with left-moving robots ('L').

Which companies ask LeetCode 2751. Robot Collisions in interviews?

This problem is frequently asked at Deutsche Bank, Samsung, Sprinklr.

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Robot Collisions - LeetCode 2751 Solution

Robot Collisions - Complete Solution Guide

Robot Collisions is LeetCode problem 2751, 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

There are n 1-indexed robots, each having a position on a line, health, and movement direction. You are given 0-indexed integer arrays positions , healths , and a string directions ( directions[i] is either 'L' for left or 'R' for right ). All integers in positions are unique . All robots start moving on the line simultaneously at the same speed in their given directions. If two robots ever share the same position while moving, they will collide . If two robots collide, the robot with lower heal

Detailed Explanation

The problem describes a scenario where robots with unique positions, health points, and movement directions ('L' for left, 'R' for right) move simultaneously on a line. When two robots collide (occupy the same position), their health points are compared. The robot with lower health is removed, and the other robot's health decreases by one. If they have equal health, both are removed. The goal is to return the health values of the surviving robots in their original order from the input.

Solution Approach

The provided solution first sorts the robots based on their initial positions to simulate the collisions correctly. Then, it uses a stack to keep track of robots moving to the right. When a robot moving to the left encounters a right-moving robot (top of the stack), they collide. The collision is resolved by comparing their health points, and the surviving robot (if any) is updated or re-pushed onto the stack if it's moving right or continues colliding to the left if it is moving left. After processing all robots, the solution filters the healths array to return only the positive health values of the surviving robots in their original order.

Step-by-Step Algorithm

Step 1: Create an array of indices and sort it based on the robots' positions.
Step 2: Initialize an empty stack to store the indices of robots moving to the right ('R').
Step 3: Iterate through the sorted indices:
Step 4: If the robot at the current index is moving to the right, push its index onto the stack.
Step 5: If the robot is moving to the left, enter a `while` loop as long as the stack is not empty and the robot's health is greater than 0. Pop the index of the right-moving robot from the stack.
Step 6: Compare the health of the left-moving robot and the right-moving robot. Update the health of both robots based on the collision rules (remove the weaker one or both if equal health). If the right-moving robot survived, push its index back onto the stack.
Step 7: After processing all robots, create a result list and add the remaining positive health values from the original healths array.
Step 8: Return the list of surviving robot healths.

Key Insights

Insight 1: Sorting by position is crucial to process collisions in the correct order as they happen on the number line.
Insight 2: A stack is an effective data structure for keeping track of robots moving right ('R') and waiting for potential collisions with left-moving robots ('L').
Insight 3: In-place modification of the healths array avoids the need for extra space to track robot health.

Complexity Analysis

Time Complexity: O(n log n)

Space Complexity: O(n)

Topics

This problem involves: Array, Stack, Sorting, Simulation.

Companies

Asked at: Deutsche Bank, Samsung, Sprinklr.

Robot Collisions | LeetCode 2751

Hard

Array Stack Sorting Simulation

Time: O(n log n)Space: O(n)

Solution Code

Available in:

import java.util.*;

class Solution {
    public List<Integer> survivedRobotsHealths(int[] positions, int[] healths, String directions) {
        int n = positions.length;
        List<Integer> indices = new ArrayList<>();
        for (int i = 0; i < n; i++) {
            indices.add(i);
        }

        indices.sort(Comparator.comparingInt(i -> positions[i]));

        Stack<Integer> stack = new Stack<>();

        for (int i : indices) {
            if (directions.charAt(i) == 'R') {
                stack.push(i);
            } else {
                int current_robot_idx = i;
                while (!stack.isEmpty() && healths[current_robot_idx] > 0) {
                    int right_robot_idx = stack.pop();

                    if (healths[current_robot_idx] > healths[right_robot_idx]) {
                        healths[current_robot_idx] -= 1;
                        healths[right_robot_idx] = 0;
                    } else if (healths[current_robot_idx] < healths[right_robot_idx]) {
                        healths[right_robot_idx] -= 1;
                        healths[current_robot_idx] = 0;
                        stack.push(right_robot_idx);
                    } else {
                        healths[current_robot_idx] = 0;
                        healths[right_robot_idx] = 0;
                    }
                }
            }
        }

        List<Integer> result = new ArrayList<>();
        for (int h : healths) {
            if (h > 0) {
                result.add(h);
            }
        }
        return result;
    }
}

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Problem Info

Description

Problem 2751: Robot Collisions - Detailed explanation and solution approaches.

Topics

Array Stack Sorting Simulation

Companies

Deutsche BankSamsungSprinklr

Stats

Difficulty:Hard

Languages:4

Companies:3

Problem #:2751

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: Create an array of indices and sort it based on the robots' positions.
2
Step 2: Initialize an empty stack to store the indices of robots moving to the right ('R').
3
Step 3: Iterate through the sorted indices:
4
Step 4: If the robot at the current index is moving to the right, push its index onto the stack.
5
Step 5: If the robot is moving to the left, enter a `while` loop as long as the stack is not empty and the robot's health is greater than 0. Pop the index of the right-moving robot from the stack.
6
Step 6: Compare the health of the left-moving robot and the right-moving robot. Update the health of both robots based on the collision rules (remove the weaker one or both if equal health). If the right-moving robot survived, push its index back onto the stack.
7
Step 7: After processing all robots, create a result list and add the remaining positive health values from the original healths array.
8
Step 8: Return the list of surviving robot healths.

💡

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 log n)

✨ Log-linear - Great for sorting algorithms!

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

Sorting the robots by position takes O(n log n) time. Iterating through the robots and simulating collisions takes O(n) time in the worst case because each robot is pushed onto the stack at most once and popped at most once. The space complexity is O(n) because the stack can potentially store all 'R' robots, and the index array also takes O(n) space. In the C version there's additional malloc/free overhead that adds slightly to execution time.

📖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

Priority Queue

Instead of sorting upfront, you could use a priority queue to maintain the order of collisions as they occur. This would involve simulating the movement of robots over time and calculating the next collision point. This approach would be significantly more complex to implement and debug.

Time: Potentially O(n^2 log n) in the worst case, depending on the number of collisions and the efficiency of the priority queue operations.Space: O(n)

Trade-offs: This is not recommended due to its complexity and likely worse performance compared to the sorting-based stack solution. The main solution is much easier to understand and implement.

✨

Key Insights

1Insight 1: Sorting by position is crucial to process collisions in the correct order as they happen on the number line.
2Insight 2: A stack is an effective data structure for keeping track of robots moving right ('R') and waiting for potential collisions with left-moving robots ('L').
3Insight 3: In-place modification of the healths array avoids the need for extra space to track robot health.

📚

Key Terms Explained

For Students

Array

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

Stack

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

Sorting

Arranging elements in a specific order (ascending/descending).

Edge Cases

Edge case 1: All robots move in the same direction. The stack will either remain empty or be filled with all robots moving right. In either case, no collisions occur, and the initial health values are returned.
Edge case 2: Collisions result in all robots being destroyed. The healths array will be modified to have all values equal to 0, and an empty array will be returned.
Edge case 3: Robots with very large health values. The solution correctly handles this as health values are integers, and the comparison logic is sound.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not sorting the robots by position before processing them. This leads to incorrect collision order and wrong results.

✗Common mistake 2: Failing to handle the case where robots have equal health values during a collision, resulting in one or both robots incorrectly surviving.

✗Common mistake 3: Not properly updating the health values after each collision, leading to inaccurate health tracking.

✗Common mistake 4: Incorrectly adding the surviving right moving robot's index back to the stack. Needs to happen only if its health is still greater than 0 after the collision.

✅

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: Simulating traffic flow with cars moving in different directions and encountering collisions.

🏦Real-world scenario 2: Modeling particle interactions in physics, where particles have different properties and can collide and annihilate each other.

💡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

Hard problems often combine multiple techniques. Break the problem into sub-problems.

⚠️ Watch Out

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

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

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.

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.

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 "Robot Collisions"?

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 stack techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

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