Race Car


Problem Classification
This problem falls under the category of simulation in combination with dynamic programming or recursive decision making. It involves simulating the movement of a car on an infinite number line according to two distinct instructions: ‘A’ (accelerate) and ‘R’ (reverse).
‘What’ Components:
 Initial condition: The car starts at position 0 and speed +1 on an infinite number line. The car can go into negative positions.
 Instructions ‘A’ (accelerate) and ‘R’ (reverse): Specific actions that can be taken. ‘A’ accelerates the car (position += speed, speed *= 2), and ‘R’ reverses the direction of the car’s speed (If the speed is positive then speed = 1, otherwise speed = 1).
 Target position: The destination that the car must reach. The goal is to reach this position using the shortest sequence of instructions.
 Output: The length of the shortest sequence of instructions to get to the target position.
This problem is a Dynamic Programming problem due to the requirement of an optimal solution (shortest sequence of instructions), which is typically achieved by making the best choice at each decision point. It also has aspects of Simulation (as it simulates the movement of a car based on instructions), and Combinatorial Optimization (as it seeks the best combination of steps to reach the target). It also contains elements of Graph Theory if you consider each state of the car (position, speed) as a node and each instruction as an edge.
The problem requires efficient decision making based on the current state of the car (position, speed) to reach the target in the least number of steps, which is characteristic of dynamic programming problems. Also, it may require exploring multiple paths and backtracking, which aligns with the principles of recursion. Hence, a good strategy would involve a blend of these computational problemsolving techniques.
Constraints
Looking at the constraints and the problem statement, we can identify a few important characteristics or conditions:
Finite target range: The target position is always a positive number and is capped at 10^4. This gives us a finite number of target positions to consider. This can potentially allow us to use dynamic programming as an approach, where we can memoize or tabulate optimal solutions for smaller subproblems (i.e., lower target positions) and then use these to build up to our final solution.
Limited actions: The car can only perform two types of actions: ‘A’ (accelerate) and ‘R’ (reverse). This simplicity restricts the number of potential decision paths.
Acceleration doubles the speed: This is an interesting condition as it means that the car’s speed grows exponentially with consecutive ‘A’ instructions. This could allow us to reach high target positions more quickly and could be a useful aspect to exploit, particularly when the target position is high. However, it also makes control of the car’s position more difficult because the position can overshoot the target very quickly.
Reverse does not change the position: The ‘R’ action only changes the speed of the car but not its position. This allows for the adjustment of the direction of the car without affecting its current position, which can be used strategically to adjust the car’s trajectory without altering its current position.
Speed and Position coupling: The speed of the car directly influences the position of the car during acceleration. This coupling can be used to plan the ‘A’ and ‘R’ actions strategically.
From the above characteristics, a plausible strategy could be to accelerate until we get close or overshoot the target, and then use the reverse instruction to finetune the position. Due to the potential to overshoot targets and because reversing doesn’t change the position, we will need to consider paths where we overshoot the target and then reverse to get back. All these aspects suggest that a depthfirst search or breadthfirst search strategy may be appropriate, where we explore all potential sequences of acceleration and reversing to find the minimum number of instructions to reach the target. The fact that the target range is capped allows for this to be feasible.
Thought Process
The problem statement seems to suggest a decisionmaking process where, at every point, there are two possible actions to take: “Accelerate” or “Reverse”. This hints towards a depthfirst search (DFS) or breadthfirst search (BFS) type of problem, where we have to traverse through all possible sequences of decisions to find the shortest one that leads to our goal.
The fact that the target and speed can both be negative implies that our search will likely need to cover both positive and negative numbers. The instructions also seem to imply that the car’s speed doubles with each acceleration, suggesting that the problem space expands rapidly with each step.
Considering the cues from the problem statement, an insightful approach might be to model this as a graph problem, where each node represents a state defined by the car’s position and speed. The goal would then be to find the shortest path from the start state to the target state.
To translate this into code, we could use BFS or DFS to traverse the graph. However, given the nature of the problem, BFS is a better fit because it naturally finds the shortest path in unweighted graphs. We also need to keep track of visited states to avoid infinite loops. Here is a Python solution:


This Python solution defines a state as a tuple of position and speed, and uses a deque to implement BFS. It also uses a set to keep track of visited states. In each iteration, it pops a state from the deque, checks if it reaches the target, and if not, adds new states to the deque. The process continues until the deque is empty or the target is reached.
Solution Approach and Analysis
This problem can be visualized as a tree problem where each node represents a state of the car. Each state is characterized by the current position and speed of the car. From each state, we can transition into two different states: one by accelerating, and the other by reversing. This creates a binary tree where each node has at most two children.
The approach can be broken down into the following steps:
Initialization: Start with an initial state where the position and speed of the car are both 0 and 1 respectively.
Tree Expansion: From the current state, generate two new states: one representing acceleration (position += speed, speed *= 2) and one representing reversal (speed = speed if speed > 0 else speed).
Goal Checking: If the current position equals the target, record the number of instructions required to reach this state (the depth of the tree). If the current position exceeds the target, we need to reverse and then accelerate until we reach the target.
Search Strategy: Repeat the steps above until we have explored all possible states or we have found a state that equals the target position. For efficiency, we can use a breadthfirst search (BFS) strategy to ensure we find the shortest sequence of instructions.
Tracking and Pruning: To avoid exploring the same state multiple times, keep a record of the visited states (position and speed) and their associated number of steps. If a state has been visited with fewer or equal steps before, we can prune this branch and not explore it further.
If we increase the target position, the number of potential states to be explored generally increases, leading to a longer runtime. However, due to the pruning step and the binary nature of the tree (accelerate or reverse), the problem is manageable within the constraints.
Example
Consider a simple case where target = 3. Start at position 0 and speed 1.
 Accelerate: position = 0 + 1 = 1, speed = 1*2 = 2.
 Accelerate again: position = 1 + 2 = 3, speed = 2*2 = 4.
So, the shortest sequence of instructions is “AA”. The car position changes as follows: 0 > 1 > 3, reaching the target position of 3.
Language Agnostic Coding Drills
 Concepts involved:
Concept 1: Understanding basic Python syntax and data structures: This is essential to read, write, and understand Python code. Includes knowing how to define functions, use control flow structures, create and use variables, and work with builtin data types like lists, tuples, sets, and dictionaries.
Concept 2: Understanding and implementing BFS: BFS or BreadthFirst Search is an algorithm for traversing or searching tree or graph data structures. It starts at the tree root (or some arbitrary node of a graph, sometimes referred to as a ‘search key’) and explores all of the neighbor nodes at the present depth prior to moving on to nodes at the next depth level.
Concept 3: Representing a problem as a graph: In many problems, it’s helpful to represent the problem space as a graph, where nodes represent states and edges represent transitions between states.
Concept 4: Managing states and transitions: For graph traversal problems, it’s often necessary to keep track of which states have been visited and what transitions are possible from each state.
 Difficulty Order:
Basic Python Syntax and Data Structures: Easy  This is the fundamental requirement for coding in Python.
Representing a problem as a graph: Medium  This requires the ability to abstract the problem and understand it from a graph theory perspective.
Understanding and implementing BFS: Medium  Understanding the BFS algorithm requires a solid understanding of trees, graphs, and queues. Implementing it requires combining this understanding with Python coding skills.
Managing states and transitions: Hard  This requires a high level of abstraction and a strong understanding of the problem space. It often involves creating and manipulating complex data structures.
 ProblemSolving Approach:
Start by understanding the problem and determining how it can be represented as a graph. Here, each state of the graph is represented by a tuple of the car’s position and speed.
Understand the transitions allowed by the problem. In this case, the car can either accelerate (which changes its position and speed) or reverse (which changes its speed only).
Implement the BFS algorithm to explore this graph. Use a queue to manage the states to be explored, and a set to keep track of already visited states.
In each step of BFS, dequeue a state, check if it is the target state, and if not, enqueue its possible next states.
Continue this process until you find the target state or have explored all reachable states. The number of steps in BFS corresponds to the minimum number of instructions to reach the target.
Targeted Drills in Python
 Coding drills for identified concepts:
 Basic Python Syntax and Data Structures:


 Representing a problem as a graph:


 Understanding and implementing BFS:


 Managing states and transitions:


 ProblemSpecific Concepts:
 Simulating car actions (A and R):


 Integrating the drills:
We start by initializing the BFS queue and the set of visited states. Each state is a tuple of the car’s current position and speed.
For each state, we dequeue it, check if it’s the target state, and if not, we simulate both possible actions (A and R), and add the resulting new states to the queue.
We keep track of the number of actions taken in BFS, which corresponds to the number of instructions.
This process continues until we find the target state or have explored all reachable states.
Q&A
The expected optimum to reach 5 is 7 steps. Maybe I haven’t understood the problem completely: I have no idea how 5 can be reached in 7 steps. The minimum I can think of is 8 (AAARARRA).
It is possible to reach the target 5 in 7 steps using the sequence “AARAARA”.
Let’s break it down:
 After the first “A”, the position is 1 (speed is 2).
 The second “A” accelerates the car to position 3 (speed is 4).
 “R” reverses the direction and sets the speed to 1 (position stays at 3).
 The first “A” after “R” moves the car back to position 2 (speed is 2).
 The second “R” reverses the direction again and sets the speed to 1 (position stays at 2).
 The first “A” after the second “R” moves the car to position 3 (speed is 2).
 The second “A” after the second “R” moves the car to the target position 5 (speed is 4).
So, the sequence “AARAARA” helps us reach the position 5 in 7 steps, which is optimal for this case.
Bfs solution is understandable. But I still didn’t get the pruning step (limiting search space to within 0 and 2*target). DP solution is not understandable: Recursive case is not understandable, overlapping state is clear (requires speed along with position rather than just position), and not sure if this is a DAG or not.
I understand that you’re finding the pruning step in the BFS solution and the recursive case in the DP solution difficult to understand. Let’s break these down.
Pruning step in BFS:
The pruning step is made based on an observation that when the position is greater than target, the car’s speed must be negative; and when the position is less than target, the car’s speed must be positive. This makes sense because if you’ve passed the target, you need to be moving back towards it (negative speed) and if you’re before the target, you need to be moving forward (positive speed).
Limiting the search space to between 0 and 2*target helps ensure we don’t do unnecessary computations for positions that are too far away from the target. The number 2 is not an absolute and could be tuned based on further analysis. However, in practice, it has been found to be a reasonable bound that allows the BFS algorithm to find the shortest path quickly.
Recursive case in DP:
In the DP solution, for a given position ‘p’ and speed ’s’, we calculate the minimum number of steps to reach the target from there. We do this by considering two actions at each point: accelerate or reverse. If we accelerate, the new position becomes ‘p + s’ and the new speed becomes ‘2 * s’. If we reverse, the position remains the same ‘p’, but the speed becomes ‘1’ if it was positive or ‘1’ if it was negative.
For each of these new positions and speeds, we recursively calculate the minimum number of steps to reach the target and take the minimum of those. The results are memoized to avoid repeated calculations for the same position and speed. This forms the recursive case for the DP solution.
This is a form of dynamic programming because we are breaking down a larger problem (minimum steps to reach the target from a given position) into smaller subproblems (minimum steps to reach the target from a closer position). The directed acyclic graph (DAG) in this case would be formed by positions as nodes and actions (accelerate or reverse) as edges, but it’s more conceptual and isn’t explicitly built in the code.
I hope this helps! Please let me know if you have further questions.
Isn’t “intuition” exactly the opposite of being “logical”? except when your can prove your intuition is correct, during interview; then shouldn’t this required proof be somewhat lightweight?
While it’s true that intuition and logic can sometimes seem to be at odds, they often work hand in hand when it comes to problemsolving, particularly in fields like programming and mathematics.
Intuition can be thought of as understanding or knowledge that comes from within, often without the need for conscious reasoning. It’s a sort of “gut feeling” or insight that you may have about a problem or situation. Logic, on the other hand, is a formal system of reasoning that follows strict rules of deduction and induction.
In the context of solving a complex problem, like in a coding interview, intuition might guide you towards a potential solution approach, based on your past experiences and the patterns you’ve recognized in similar problems. Once you have that intuitive idea, you would then use logic to rigorously test it, flesh it out into a full solution, and ensure that it works for all edge cases. So in this sense, intuition can provide the initial spark of insight, while logic carries it to the finish line.
Regarding your question about proving your intuition during an interview  it’s always a good idea to explain your thought process to the interviewer. Even if your intuition is guiding you towards a solution, try to articulate the reasons behind your hunch. The interviewer is interested in understanding how you approach and solve problems, and a key part of this is being able to communicate your ideas clearly.
It’s also worth noting that while proving your intuition might require rigorous logical reasoning, the key is to break down the problem and your solution approach into simpler parts that can be more easily explained and understood. This way, the proof becomes “lightweight” in terms of its complexity and the effort required to understand it.
I have no idea about what int targ2 = walk  targ1; is? Should Dijkstra be targ2 = walk + targ1. If we start from k = 0; then targ2 = 1 and steps is zero. What does 4 for 0 step meaning? And I am also confuse about Math.floorMod? I think it is just for mapping. I am very curious about how it works and has been thinking for one day.
The variable targ2
is defined as walk  targ1
because it is used to keep track of the remaining distance to the target that needs to be covered. In the context of the problem, walk
is the total distance covered by the car, and targ1
is the part of this distance that is already considered in the calculation. Therefore, targ2
represents the remaining part of walk
that still needs to be considered. If you set targ2
as walk + targ1
, it would instead represent the total distance that has been considered twice, which is not what is intended in this context.
As for the usage of Math.floorMod
, this method in Java returns the floor modulus of the two arguments, which is a signed remainder. It is indeed used for mapping in this context, specifically for mapping the remainder of targ2
divided by 2*walk
. This helps ensure that the calculated steps always stay within the bounds of the array.
Lastly, the 4
for 0 steps is used as a placeholder or initial value. It essentially represents a “nonvalue” or an “invalid value”, which gets updated in the calculation process. This is a common practice in programming when you need to initialize a variable but don’t have a valid or meaningful value to assign to it at the time of initialization.
Why is it ‘Math.abs(nextPosition target) < target’ ?
This condition Math.abs(nextPosition  target) < target
is a pruning step in the algorithm.
The expression Math.abs(nextPosition  target)
represents the absolute difference between the car’s new position and the target. If this difference is greater than or equal to target
, it means the car has moved an equal or greater distance away from the target than its initial position. This means that this movement sequence is definitely not optimal because the car has moved farther away from the target than its starting position.
In such cases, there’s no need to explore this sequence any further because we already know it’s not going to give us an optimal result. Therefore, this condition is used to limit the search space and improve the efficiency of the solution by eliminating suboptimal paths early in the search process.
Is it possible to reverse twice like AAARRA?
Based on the problem statement, the car can indeed reverse twice (or any number of times for that matter). Each “R” command simply reverses the direction of the car. So “AAARRA” would mean:
 “AA”: Accelerate twice, moving the car to position 3 (0 > 1 > 3).
 “RR”: Reverse twice, which effectively leaves the car’s direction unchanged but its speed resets to 1.
 “A”: Accelerate once more, moving the car to position 4 (3 > 4).
In this scenario, the double reversal (“RR”) effectively acts as a brake, stopping the car’s acceleration and resetting its speed to 1 without changing its direction. This could be beneficial in situations where you want to gain fine control over the car’s position without necessarily wanting to reverse direction.
I had “AAARAA”, but the answer is 5
The problem statement indicates that the goal is to reach a target position using the fewest number of instructions. Given a target position, we need to find the sequence of instructions (‘A’ for accelerate and ‘R’ for reverse) that gets us to the target position using the minimum number of instructions.
If your target was 4, the sequence “AAARAA” would indeed get you to the target, but it’s not the shortest sequence of instructions.
To reach a position 4, you can follow these steps:
 “A”: Accelerate once, moving the car to position 1.
 “A”: Accelerate again, moving the car to position 3.
 “RA”: Reverse and then Accelerate, moving the car back to position 2, and then to position 4.
So, the sequence “AARA” gets you to position 4 using only 4 instructions. It’s important to note that the sequence “AAARAA” also gets you to the same position, but it uses 6 instructions, which is not the shortest sequence.
Therefore, for target = 4, the minimum number of instructions is indeed 4, not 6.
I just can’t figure out the anwser at all. As far I can derive, you need at least 8 steps to reach 5. Can anybody show me the actions list?
Sure, I can show you how to reach the position 5 with 7 steps, which is the minimum number of steps. Here’s the sequence of actions:
 “A”: Accelerate once, moving the car to position 1.
 “A”: Accelerate again, moving the car to position 3.
 “A”: Accelerate a third time, moving the car to position 7.
 “R”: Reverse, but the position stays the same.
 “A”: Accelerate, but now in reverse, moving the car back to position 6.
 “R”: Reverse again, but the position stays the same.
 “A”: Accelerate, moving the car forward to position 5.
So, the sequence “AAARARA” gets you to position 5 using 7 instructions. Please note that this problem is rather complex and requires a good understanding of dynamic programming to be solved efficiently.
Check how to reach 5. The test case shows the answer is 7. But think it seriously, can we really reach 5 with 7 operation?
case 1: if we reach at 7, this takes 3. And we would like to get back. Now for the first time, it takes RA as 2. Now our speed is 2, and we reach at 6. Can we reach at 5 with another 2 moves? No! We need another 3 moves, RRA to reach at 5. Because our speed is 2! So the final cost is 8. case 2: we reach at 3. It takes 2. Now we clear speed and move 1. It takes RRA. And now our speed is 2. So we need to clear our speed by RRA again. So the final cost 8.
Find something interesting? Why most answer gives 7 and the test case is 7? Sometimes we arrives with a negative speed, and sometimes it is a positive speed. We cannot dirrectly use that result. We need to check the direction before we use it! But almost case assumes the speed is positive! And that would mislead the final result. In the 5 case, we arrives 6 with 5. And now we believe we could arrive 5 with another RA. That would never work, because now we have a neagtive speed. So, for the dp. It must follows with a boolean status to show which direction is not the speed it is. And for the same direction, we must use RR to reset speed, not R.
You’re right in your analysis. The dynamic programming solution is more nuanced than simply finding the smallest number of steps to reach a specific position. The direction in which the car is facing (indicated by whether the speed is positive or negative) plays a crucial role. This is why the state in DP is represented as (position, speed), instead of just position.
When arriving at a certain position, the speed can be either positive or negative. If the speed is positive, it’s possible to continue moving forward without reversing the direction. But if the speed is negative, the direction must be reversed (using ‘R’) to continue moving towards the target. This costs an additional step.
The sequence of commands to reach position 5 is indeed “AAARARA” which consists of 7 commands, not 8. Here is how it goes:
 “A”: Accelerate once, moving the car to position 1.
 “A”: Accelerate again, moving the car to position 3.
 “A”: Accelerate a third time, moving the car to position 7.
 “R”: Reverse, the position stays at 7.
 “A”: Accelerate, but now in reverse, moving the car back to position 6.
 “R”: Reverse again, the position stays at 6.
 “A”: Accelerate, moving the car forward to position 5.
So, in the end, the car arrives at position 5, with a total of 7 commands, and a speed of 1. Therefore, the direction is right for the car to continue moving towards larger positions if needed. This sequence of commands is the minimum possible to reach the position 5, hence the answer is 7.
Trying make sense of why we add +1 in finding the minimum value to update the dp[] array
The reason for adding 1 is to account for the “R” (reverse) operation. The variables forward
and backward
in this code represent the number of steps or the number of “A” (accelerate) operations.
In the first case where forwardDistance > i
, the car has overshot the target i
and needs to go back by forwardDistance  i
. The number of “A” operations here is forward
but you also have to add 1 for the “R” operation that was used to change the direction of the car.
So forward + 1 + dp[forwardDistance  i]
actually means: number of forward operations (or “A” operations) + one reverse operation (or “R” operation) + the number of operations needed to reach the target from the current position.
In the second case where forwardDistance < i
, after doing some forward operations and one reverse operation, it also does some backward operations (which includes another reverse operation) to reach the target.
So forward + 1 + backward + 1 + dp[i  forwardDistance + backwardDistance]
means: number of forward operations + one reverse operation + number of backward operations + another reverse operation + the number of operations needed to reach the target from the current position.
Therefore, the “+1” is used to account for the “R” (reverse) operation in both these cases.