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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
    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

Solutions

Solution 1

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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]

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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];
    }
}

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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];
    }
};

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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]
}

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