818. Race Car - Detailed Explanation

Problem Statement

Description:
You start at position 0 with a speed of +1 on an infinite number line. You have two available commands:

• Accelerate ("A"):

  • Position becomes: position += speed
  • Speed becomes: speed *= 2

• Reverse ("R"):

  • If the current speed is positive, set speed to -1.
  • If the current speed is negative, set speed to +1.
  • Position does not change.

Given a target position (a positive integer), your goal is to reach exactly that position using the minimum number of commands.

Examples:

• Example 1:

  • Input: target = 3
  • Output: 2
  • Explanation:
    One optimal sequence is:
    • Command 1: "A" → position = 0 + 1 = 1, speed = 2
    • Command 2: "A" → position = 1 + 2 = 3, speed = 4

• Example 2:

  • Input: target = 6
  • Output: 5
  • Explanation:
    One optimal sequence is:
    • "A": position = 1, speed = 2
    • "A": position = 3, speed = 4
    • "R": reverse to speed = -1 (position remains 3)
    • "A": position = 3 + (-1) = 2, speed = -2
    • "A": position = 2 + (-2) = 0, speed = -4
      (Note: This is one sequence; there exists a better sequence that reaches 6 in 5 moves. The key idea is that sometimes overshooting and then reversing is optimal.)

• Example 3:

  • Input: target = 4
  • Output: 5
  • Explanation:
    The optimal sequence might involve overshooting past 4 and then using reversals to come back, resulting in a minimum command count of 5.

Constraints:

• 1 <= target <= 10^4 (The target is a positive integer.)

2. Hints Before You Start

• Hint 1: Think about the effect of repeatedly accelerating. After ( m ) consecutive "A" commands, your position becomes ( 2^m - 1 ). Use this to decide when you overshoot the target.
• Hint 2: Consider that sometimes it is optimal to overshoot the target and then reverse, while other times it may be better to reverse before overshooting. Tracking the last few moves with a recursive or dynamic programming approach is key.

Approaches

Brute Force / State-Space Search (BFS)

Idea:

• Model each state as a pair ((\text{position}, \text{speed})).
• Use Breadth-First Search (BFS) to simulate all possible sequences of commands.
• Stop when you reach the target position.

Drawbacks:

• The number of states grows quickly. Although BFS guarantees the shortest sequence, it is too slow when the target is large.

Optimal Approach (Dynamic Programming with Recurrence)

Idea:

• Let ( dp[t] ) represent the minimum number of commands needed to reach position ( t ).
• For a given target ( t ), determine the minimum number ( m ) of consecutive accelerations such that ( 2^m - 1 \geq t ).
• Case 1: If ( t = 2^m - 1 ), then the answer is simply ( m ) (because you can reach ( t ) by accelerating ( m ) times).
• Case 2: Otherwise, you have two choices:
  1. Overshoot and Reverse:
     Accelerate ( m ) times to reach ( 2^m - 1 ) (which is greater than ( t )), then reverse, and solve the subproblem for the remaining distance:
     [ dp[t] = m + 1 + dp[(2^m - 1) - t] ]
  2. Reverse Before Overshooting:
     Instead of ( m ) accelerations, try accelerating for ( m - 1 ) steps and then make a reverse. After reversing, you can perform a series of ( j ) accelerations in the opposite direction (where ( 0 \leq j < m - 1 )). This gives another candidate sequence: [ dp[t] = \min_{0 \leq j < m-1} \{ (m - 1) + 1 + j + 1 + dp[t - (2^{m-1} - 2^j)] \} ] Here, ( (m - 1) ) is the number of “A” commands to get close to ( t ), the first "R" reverses the car, then ( j ) “A” commands in reverse, another "R" to face the target again, and finally solve the remaining distance.
• Use recursion with memoization (or bottom-up DP) to compute ( dp[t] ) for every ( t ) up to the target.

Visual Example (Simplified Concept):

Suppose ( t = 4 ):

• Find ( m ) such that ( 2^m - 1 \geq 4 ). Here, ( m = 3 ) because ( 2^3 - 1 = 7 ).
• Since ( 7 \neq 4 ), try:

  • Overshoot Case:
    Use 3 "A" commands (reaching 7), then "R", then solve for ( 7 - 4 = 3 ).

  • Reverse Before Overshoot:
    Try accelerating for ( m-1 = 2 ) commands (position ( 3 )), then "R".
    Now try different values of ( j ) for reversing commands and then solve for the remaining distance.

The recurrence chooses the option with the minimum total commands.

Complexity Analysis:

• Time Complexity: Although the recurrence appears complex, the overlapping subproblems and memoization lead to an efficient solution for ( t ) up to ( 10^4 ).
• Space Complexity: O(t) for the memoization table.

Code Implementations

Python Code (Optimal DP Approach)