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Race Car  Leetcode Solution
LeetCode: Race Car Leetcode Solution
Difficulty: Hard
Topics: dynamicprogramming
The Race Car problem on LeetCode is a dynamic programming problem that requires finding the minimum number of steps (in terms of acceleration and reverse gear) required to reach a target distance. Additionally, the car can only move in the positive x direction.
Let's understand the problem statement in detail. The problem states that you are driving a car of the speed zero on an infinite 1dimensional plane. The car has acceleration and reverse motions which allow you to increase or decrease your speed at will. The aim is to reach a specific point in the least number of moves.
To solve this problem, we can break it down into smaller parts and find the optimal solution for each of them using dynamic programming. We can define an array dp
where dp[i]
represents the minimum number of steps it takes to reach distance i
.
Now, let's try to find the minimum steps required to reach a target using various cases:

If target is the power of 2, i.e.,
target = 2^k
: In this case, we have two options, either move the entire distance with one acceleration or, make a reverse move to slow down, and then accelerate to reach the target. Therefore, the optimal solution would bedp[i] = min(2^k + 1 + dp[i  (2^k  1)])
. Note that we add1
here since we need to make one more extra move after we reach the target distance. 
If target is not the power of 2: Here, we need to apply an acceleration or reverse gear to the car and find the best possible solution. Therefore, we will make
j
acceleration moves and then1
reverse move to reduce the speed. In this case, the optimal solution would bedp[i] = min(dp[i], j + 1 + dp[i  (2^k  1) + (2^(k1)  1)  (2^(kj1)  1)])
. Here,(2^k1)
represents the nearest distance to the power of 2,(2^(k1)1)
represents the distance we move in reverse gear, and(2^(kj1)1)
represents the distance we move in reverse gear after making j accelerations.
Here is the complete Python implementation:
class Solution:
def racecar(self, target: int) > int:
dp = [float('inf')] * (target + 1)
dp[0] = 0
for i in range(1, target+1):
k = i.bit_length()
if i == 2**k  1:
dp[i] = k
continue
for j in range(k1):
dp[i] = min(dp[i], (kj1) + 1 + dp[i  (2**(k1)  2**(j))])
if 2**k  1  i < i:
dp[i] = min(dp[i], k + 1 + dp[2**k  1  i])
return dp[target]
This implementation uses float('inf')
to represent an impossible distance. We initialize the dp
array with float('inf')
except for the dp[0]
which represents the minimum number of steps needed to reach a zero distance which is 0
.
The for
loop runs from 1
to target
and calculates the minimum steps needed to reach the current distance. We calculate the value of k
which represents the nearest power of 2
greater than i
.
We then check if the current distance is equal to 2**k 1
. If yes, then the optimal solution is k
moves (just move the whole distance with an acceleration move).
If the current distance is not the power of 2, then we make j
acceleration moves and then 1
reverse move to cover the distance. We use the formula mentioned above to calculate the optimal solution.
After finding the solution for acceleration moves, we check if it would be better to make a reverse move and reach the target from the other end. dp[2**k  1  i]
calculates the solution for reaching the target from the other end. We choose the minimum of the two solutions and return the result.
This solution provides the correct answer for the Race Car problem on LeetCode.
Race Car Solution Code
1