LeetCode 1824. Minimum Sideway Jumps

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: 0
Explanation: There are no obstacles on lane 2. No side jumps are required.

Example 3:

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

Constraints:

• obstacles.length == n + 1

• 1 <= n <= 5 * 10^5

• 0 <= obstacles[i] <= 3

• obstacles[0] == obstacles[n] == 0

Solution

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中文版解答Bilibili Link

class Solution {
public:
    int minSideJumps(vector<int>& obstacles) {
        int n = obstacles.size();
        vector<vector<int>> dp (n, vector<int>(3, 99999999));
        dp[0][1] = 0;
        dp[0][0] = dp[0][2] = 1;
        
        for (int i = 1; i < n; i++) {
            for (int k = 0; k < 3; k++) {
                if (obstacles[i - 1] == (k + 1) || obstacles[i] == (k + 1)) {
                    dp[i][k] = 99999999;
                } else {
                    int a_lane = (k + 1) % 3;
                    int b_lane = (k + 2) % 3;
                    dp[i][k] = min(dp[i - 1][k], min(dp[i - 1][a_lane], dp[i - 1][b_lane]) + 1);
                }
            }
        }
        
        return min(dp[n-1][0], min(dp[n-1][1], dp[n-1][2]));
    }
};