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Race Car

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", your car goes to positions 0 --> 1 --> 3 --> 3, and your speed goes to 1 --> 2 --> 4 --> -1.

Given a target position target, return the length of the shortest sequence of instructions to get there.

 

Example 1:

Input: target = 3
Output: 2
Explanation: 
The shortest instruction sequence is "AA".
Your position goes from 0 --> 1 --> 3.

Example 2:

Input: target = 6
Output: 5
Explanation: 
The shortest instruction sequence is "AAARA".
Your position goes from 0 --> 1 --> 3 --> 7 --> 7 --> 6.

 

Constraints:

  • 1 <= target <= 104

Solution Explanation for Race Car Problem

The Race Car problem involves finding the shortest sequence of instructions ('A' for accelerate and 'R' for reverse) to reach a target position starting from position 0 with speed +1. The challenge lies in the non-linear nature of the car's movement.

This solution employs dynamic programming to efficiently solve the problem. Let's break down the approach and code:

Approach

  1. DP Table: We use a dp array (or equivalent in different languages) of size target + 1. dp[i] stores the minimum number of instructions needed to reach position i.

  2. Base Cases: dp[0] = 0 (already at the target).

  3. Iteration: We iterate from i = 1 to target. For each i, we find the shortest path:

    • Optimal Acceleration: We determine k, the smallest integer such that 2<sup>k</sup> - 1 >= i. If i == 2<sup>k</sup> - 1, then we can reach i by simply accelerating k times (dp[i] = k).

    • Reverse and Continue: Otherwise, we can reach i by first accelerating to 2<sup>k</sup> - 1 (which takes k steps), then reversing (+1 step), and then going the remaining distance (2<sup>k</sup> - 1 - i). This gives us one possible solution dp[i] = dp[2<sup>k</sup> - 1 - i] + k + 1 (we use 2<sup>k</sup> - 1 - i as the remaining position in negative direction).

    • Intermediate Reversals: We can explore reaching the target through intermediate reversals. This is where we loop through possible values for j. We go to position 2<sup>k-1</sup> - 2<sup>j</sup> and then take another move. We also account for the additional instructions required. The formula accounts for the steps taken to reach the 2<sup>k-1</sup> - 2<sup>j</sup> position, the reversal, and the remaining distance. This is added to the already calculated optimal value for the remaining distance to create dp[i] = min(dp[i], dp[i - (2**(k - 1) - 2**j)] + k - 1 + j + 2)

  4. Result: dp[target] contains the minimum number of instructions to reach the target position.

Time and Space Complexity

  • Time Complexity: O(target * log(target)). The outer loop iterates up to target, and the inner loop iterates up to log(target) times (due to the k variable).
  • Space Complexity: O(target). We use an array of size target + 1 to store the DP results.

Code Explanations (with minor variations based on language)

The provided code in Python, Java, C++, and Go implements the dynamic programming approach described above. They all share the same core logic:

  1. Initialization: A DP array is created and initialized.
  2. Iteration: The code iterates to find the shortest path for each position. k is calculated using bit manipulation (bit_length() in Python, Integer.numberOfLeadingZeros() in Java, __builtin_clz() in C++, bits.Len() in Go). These functions efficiently determine the number of bits needed to represent the number which helps compute 2k.
  3. Optimal Substructure: The code effectively utilizes the optimal solutions for smaller subproblems (positions before the current i) to calculate the optimal solution for i. The min() function (or equivalent) ensures that the shortest path is always selected.
  4. Result: The final answer is the value at dp[target].

The minor variations between the languages primarily relate to syntax and built-in functions used for bit manipulation and finding the minimum value. The core algorithm remains consistent across all implementations.