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 * 105
  • 0 <= obstacles[i] <= 3
  • obstacles[0] == obstacles[n] == 0

Rust Solution

struct Solution;

impl Solution {
    fn min_side_jumps(obstacles: Vec<i32>) -> i32 {
        let n = obstacles.len();
        let mut dp: Vec<Vec<i32>> = vec![vec![0, 0, 0]; n];
        dp[0][0] = 1;
        dp[0][2] = 1;
        for i in 1..n {
            for j in 0..3 {
                if obstacles[i] == (j + 1) as i32 {
                    dp[i][j] = std::i32::MAX;
                } else {
                    let mut min = std::i32::MAX;
                    for k in 0..3 {
                        if !(obstacles[i - 1] == (k + 1) as i32 || obstacles[i] == (k + 1) as i32) {
                            min = min.min(dp[i - 1][k] + if k == j { 0 } else { 1 });
                        }
                    }
                    dp[i][j] = min;
                }
            }
        }
        *dp[n - 1].iter().min().unwrap()
    }
}

#[test]
fn test() {
    let obstacles = vec![0, 1, 2, 3, 0];
    let res = 2;
    assert_eq!(Solution::min_side_jumps(obstacles), res);
    let obstacles = vec![0, 1, 1, 3, 3, 0];
    let res = 0;
    assert_eq!(Solution::min_side_jumps(obstacles), res);
    let obstacles = vec![0, 2, 1, 0, 3, 0];
    let res = 2;
    assert_eq!(Solution::min_side_jumps(obstacles), res);
}

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