動機

oh Floyd-Warshall，我忘了!! 原來還可以這樣解loop

Problem

There is a 3 lane road of length n that consists of n + 1 points labeled from 0 to n. A frog starts at point 0 in the second lane and wants to jump to point n. However, there could be obstacles along the way.

You are given an array obstacles of length n + 1 where each obstacles[i] (ranging from 0 to 3) describes an obstacle on the lane obstacles[i] at point i. If obstacles[i] == 0, there are no obstacles at point i. There will be at most one obstacle in the 3 lanes at each point.

  • For example, if obstacles[2] == 1, then there is an obstacle on lane 1 at point 2.

The frog can only travel from point i to point i + 1 on the same lane if there is not an obstacle on the lane at point i + 1. To avoid obstacles, the frog can also perform a side jump to jump to another lane (even if they are not adjacent) at the same point if there is no obstacle on the new lane.

  • For example, the frog can jump from lane 3 at point 3 to lane 1 at point 3.

Return the minimum number of side jumps the frog needs to reach any lane at point n starting from lane 2 at point 0.

Note: There will be no obstacles on points 0 and n.

 

Example 1:

Input: obstacles = [0,1,2,3,0]Output: 2 Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps (red arrows).Note that the frog can jump over obstacles only when making side jumps (as shown at point 2).

Example 2:

Input: obstacles = [0,1,1,3,3,0]Output: 0Explanation: There are no obstacles on lane 2. No side jumps are required.

Example 3:

Input: obstacles = [0,2,1,0,3,0]Output: 2Explanation: The optimal solution is shown by the arrows above. There are 2 side jumps.

 

Constraints:

  • obstacles.length == n + 1
  • 1 <= n <= 5 * 105
  • 0 <= obstacles[i] <= 3
  • obstacles[0] == obstacles[n] == 0

sol

原本想用熟悉的方法解dp，之後就遇到很尷尬的狀態，對於不同障礙的分佈組合會處理到死

但這是path，我們有relax!! 所以可以一直relax!!

class Solution:
    def minSideJumps(self, obstacles: List[int]) -> int:
        dp = [float("inf"), 0, float("inf")]
        for o in obstacles:
            o -= 1
            if o != -1:
                dp[o] = float("inf")
            for l in range(3):
                if l != o:
                    dp[l] = min(dp[l], dp[(l+1) % 2]+1, dp[(l+2) % 3]+1)
                    # walk multiple times!!
            print(f'{dp}')
        return min(dp)