818. Race Car

Description

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

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

Solutions

• class Solution {
public int racecar(int target) {
int[] dp = new int[target + 1];
for (int i = 1; i <= target; ++i) {
int k = 32 - Integer.numberOfLeadingZeros(i);
if (i == (1 << k) - 1) {
dp[i] = k;
continue;
}
dp[i] = dp[(1 << k) - 1 - i] + k + 1;
for (int j = 0; j < k; ++j) {
dp[i] = Math.min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
}
}
return dp[target];
}
}

• class Solution {
public:
int racecar(int target) {
vector<int> dp(target + 1);
for (int i = 1; i <= target; ++i) {
int k = 32 - __builtin_clz(i);
if (i == (1 << k) - 1) {
dp[i] = k;
continue;
}
dp[i] = dp[(1 << k) - 1 - i] + k + 1;
for (int j = 0; j < k; ++j) {
dp[i] = min(dp[i], dp[i - (1 << (k - 1)) + (1 << j)] + k - 1 + j + 2);
}
}
return dp[target];
}
};

• class Solution:
def racecar(self, target: int) -> int:
dp = [0] * (target + 1)
for i in range(1, target + 1):
k = i.bit_length()
if i == 2**k - 1:
dp[i] = k
continue
dp[i] = dp[2**k - 1 - i] + k + 1
for j in range(k - 1):
dp[i] = min(dp[i], dp[i - (2 ** (k - 1) - 2**j)] + k - 1 + j + 2)
return dp[target]


• func racecar(target int) int {
dp := make([]int, target+1)
for i := 1; i <= target; i++ {
k := bits.Len(uint(i))
if i == (1<<k)-1 {
dp[i] = k
continue
}
dp[i] = dp[(1<<k)-1-i] + k + 1
for j := 0; j < k; j++ {
dp[i] = min(dp[i], dp[i-(1<<(k-1))+(1<<j)]+k-1+j+2)
}
}
return dp[target]
}