What is the time complexity of Race Car?

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

What data structures are used to solve Race Car?

This problem uses Dynamic Programming. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Race Car?

Insight 1: Dynamic programming is suitable because the optimal path to a given position can be built from the optimal paths to previous positions. Insight 2: The key is to consider two main strategies: overshooting the target and reversing, or undershooting the target and maneuvering to reach it.

Which companies ask LeetCode 818. Race Car in interviews?

This problem is frequently asked at Anduril, Turing.

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Race Car - LeetCode 818 Solution

Q: What is the space complexity of Race Car?

The space complexity is O(target).

Race Car - Complete Solution Guide

Race Car is LeetCode problem 818, 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

Your car starts at position 0 and speed +1 on an infinite number line. Your car can go into negative positions. Your car drives automatically according to a sequence of instructions 'A' (accelerate) and 'R' (reverse): When you get an instruction 'A' , your car does the following: position += speed speed *= 2 When you get an instruction 'R' , your car does the following: If your speed is positive then speed = -1 otherwise speed = 1 Your position stays the same. For example, after commands "AAR" ,

Detailed Explanation

The "Race Car" problem presents a scenario where a car starts at position 0 with a speed of +1 on an infinite number line. The car can only perform two actions: 'A' (accelerate) and 'R' (reverse). Accelerate increases the position by the current speed and doubles the speed. Reverse changes the speed to -1 if positive or +1 if negative, while the position remains unchanged. The goal is to find the shortest sequence of 'A' and 'R' instructions to reach a given target position.

Solution Approach

The provided solution utilizes dynamic programming to find the minimum number of instructions needed to reach each position from 0 to the target. It builds a DP table where `dp[i]` stores the minimum number of instructions to reach position `i`. The algorithm iterates from 1 to the target, calculating `dp[i]` based on two possible strategies: 1. Overshoot the target with a sequence of 'A' instructions and then reverse and come back. 2. Undershoot the target with a shorter sequence of 'A' instructions, reverse, and maneuver to reach the target. The algorithm compares the number of steps required by each strategy and chooses the minimum.

Step-by-Step Algorithm

Step 1: Initialize a DP table `dp` of size `target + 1` with all values set to 0. `dp[i]` will store the minimum instructions to reach position `i`.
Step 2: Iterate from `i = 1` to `target` to populate the DP table.
Step 3: For each position `i`, calculate `k`, the number of 'A' instructions to overshoot the target or reach a point close to it. `k` is the number of bits required to represent `i + 1`. This is equivalent to `ceil(log2(i + 1))` or `i.bit_length()` in Python.
Step 4: If `i == (1 << k) - 1`, meaning the car can reach `i` by simply performing `k` 'A' instructions, then `dp[i] = k`.
Step 5: Otherwise, calculate the number of instructions using the two strategies:
Step 6: Overshoot strategy: Perform `k` 'A' instructions (reaching position `(1 << k) - 1`), one 'R' instruction, and then use the optimal number of instructions to reach the remaining distance `(1 << k) - 1 - i`. The total steps are `k + 1 + dp[(1 << k) - 1 - i]`.
Step 7: Undershoot strategy: Perform `k - 1` 'A' instructions, one 'R' instruction, then `j` 'A' instructions (where `j` ranges from `0` to `k - 2`), one 'R' instruction and then calculate the optimal path to the remaining position.
Step 8: Calculate position after maneuver as `(1 << (k - 1)) - (1 << j)`. Then calculate the number of steps as `(k-1) + 1 + j + 1 + dp[i - pos_after_maneuver]`
Step 9: Take the minimum steps between overshoot and undershoot.
Step 10: Set `dp[i]` to the minimum number of instructions calculated in step 9.
Step 11: After the loop finishes, `dp[target]` contains the minimum number of instructions to reach the target.

Key Insights

Insight 1: Dynamic programming is suitable because the optimal path to a given position can be built from the optimal paths to previous positions.
Insight 2: The key is to consider two main strategies: overshooting the target and reversing, or undershooting the target and maneuvering to reach it.
Insight 3: Leveraging bit manipulation (specifically bit length) helps determine the number of accelerations needed to overshoot or undershoot the target.

Complexity Analysis

Time Complexity: O(target*log(target))

Space Complexity: O(target)

Topics

This problem involves: Dynamic Programming.

Companies

Asked at: Anduril, Turing.

Race Car | LeetCode 818

Hard

Dynamic Programming

Time: O(target*log(target))Space: O(target)

Solution Code

Available in:

class Solution {
    public int racecar(int target) {
        int[] dp = new int[target + 1];
        for (int i = 1; i <= target; i++) {
            int k = Integer.toBinaryString(i).length();

            if (i == (1 << k) - 1) {
                dp[i] = k;
                continue;
            }

            // Option 1: Overshoot with k 'A's and come back
            int res = k + 1 + dp[(1 << k) - 1 - i];

            // Option 2: Undershoot with k-1 'A's, then maneuver
            for (int j = 0; j < k - 1; j++) {
                int pos_after_maneuver = (1 << (k - 1)) - (1 << j);
                int steps = (k - 1) + 1 + j + 1 + dp[i - pos_after_maneuver];
                res = Math.min(res, steps);
            }

            dp[i] = res;

        }
        return dp[target];
    }
}

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Problem Info

Description

Problem 818: Race Car - Detailed explanation and solution approaches.

Topics

Dynamic Programming

Companies

AndurilTuring

Stats

Difficulty:Hard

Languages:4

Companies:2

Problem #:818

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

11 steps

1
Step 1: Initialize a DP table `dp` of size `target + 1` with all values set to 0. `dp[i]` will store the minimum instructions to reach position `i`.
2
Step 2: Iterate from `i = 1` to `target` to populate the DP table.
3
Step 3: For each position `i`, calculate `k`, the number of 'A' instructions to overshoot the target or reach a point close to it. `k` is the number of bits required to represent `i + 1`. This is equivalent to `ceil(log2(i + 1))` or `i.bit_length()` in Python.
4
Step 4: If `i == (1 << k) - 1`, meaning the car can reach `i` by simply performing `k` 'A' instructions, then `dp[i] = k`.
5
Step 5: Otherwise, calculate the number of instructions using the two strategies:
6
Step 6: Overshoot strategy: Perform `k` 'A' instructions (reaching position `(1 << k) - 1`), one 'R' instruction, and then use the optimal number of instructions to reach the remaining distance `(1 << k) - 1 - i`. The total steps are `k + 1 + dp[(1 << k) - 1 - i]`.
7
Step 7: Undershoot strategy: Perform `k - 1` 'A' instructions, one 'R' instruction, then `j` 'A' instructions (where `j` ranges from `0` to `k - 2`), one 'R' instruction and then calculate the optimal path to the remaining position.
8
Step 8: Calculate position after maneuver as `(1 << (k - 1)) - (1 << j)`. Then calculate the number of steps as `(k-1) + 1 + j + 1 + dp[i - pos_after_maneuver]`
9
Step 9: Take the minimum steps between overshoot and undershoot.
10
Step 10: Set `dp[i]` to the minimum number of instructions calculated in step 9.
11
Step 11: After the loop finishes, `dp[target]` contains the minimum number of instructions to reach the target.

💡

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(target*log(target))

💾

Space Complexity

O(target)

📝

Why?

The algorithm iterates from 1 to `target` in the outer loop. Inside the loop, it calculates `k` (log(target)). The inner loop for the undershoot strategy iterates at most `k - 1` times, so this loop takes O(log(target)) time for each value i from 1 to target. The dp array of size O(target) stores previously calculated results which are used in constant time lookups within each of those iterations. The time complexity is therefore O(target * log(target)). The space complexity is O(target) because the DP table `dp` stores values for all positions from 0 to `target`.

📖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

Breadth-First Search (BFS)

BFS can be used to explore the possible sequences of instructions. Maintain a queue of (position, speed, instruction_count) tuples. Expand each tuple by applying 'A' and 'R' instructions and adding the resulting states to the queue. Keep track of visited states to avoid cycles. The first time the target is reached, the instruction_count represents the shortest sequence length.

Time: O(2^n), where n is the number of instructions. Can be very slow.Space: O(2^n)

Trade-offs: BFS can be less efficient because the number of possible instruction sequences can grow exponentially. It may be more intuitive to understand but may exceed time limit constraints for larger targets.

✨

Key Insights

1Insight 1: Dynamic programming is suitable because the optimal path to a given position can be built from the optimal paths to previous positions.
2Insight 2: The key is to consider two main strategies: overshooting the target and reversing, or undershooting the target and maneuvering to reach it.
3Insight 3: Leveraging bit manipulation (specifically bit length) helps determine the number of accelerations needed to overshoot or undershoot the target.

📚

Key Terms Explained

For Students

Dynamic Programming

Breaking complex problems into smaller subproblems and storing results.

Edge Cases

Edge case 1: target = 1. The algorithm correctly calculates the minimum number of instructions as 1 ('A').
Edge case 2: target = 2. The algorithm correctly calculates the minimum number of instructions as 3 ('AAR').

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Failing to consider both overshoot and undershoot strategies for reaching the target.

✗Common mistake 2: Not correctly handling the reversal of speed when the 'R' instruction is encountered.

✗Common mistake 3: Forgetting to consider the cost of the 'R' instruction itself when calculating the total number of instructions.

✅

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 path planning, where a robot needs to navigate to a specific location with constraints on its movement capabilities (like acceleration and deceleration).

🏦Real-world scenario 2: Game development, where an AI agent needs to find the shortest path to a target location with similar movement constraints.

💡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

Start with recursion, then memoize. The hardest part is identifying the state. Ask yourself: "What information do I need to make the current decision?"

🎯 Interview Tip

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

⚠️ Watch Out

Forgetting base cases or initializing the DP array incorrectly. Start by solving the smallest subproblems.

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: September 7, 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 13

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

@TechLeadJul 6

❓

Frequently Asked Questions

How do I identify if a problem needs dynamic programming?

DP problems typically have: 1) Overlapping subproblems (same calculation repeated), 2) Optimal substructure (optimal solution uses optimal solutions to subproblems). Look for keywords like "minimum/maximum," "count ways," "is it possible," or "longest/shortest."

How do I approach hard algorithm problems?

Hard problems often require: 1) Breaking down into subproblems, 2) Combining multiple advanced techniques, 3) Careful optimization of brute force. Start with a working solution (even O(n²) or worse), then optimize. Draw examples, identify patterns, and think about what makes this problem unique.

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 "Race Car"?

For this hard-level Dynamic Programming 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.

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