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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

• Java
• C++
• Python
• Go

• 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]
  }
  

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