1091. Shortest Path in Binary Matrix

In an N by N square grid, each cell is either empty (0) or blocked (1).

A clear path from top-left to bottom-right has length k if and only if it is composed of cells C_1, C_2, ..., C_k such that:

  • Adjacent cells C_i and C_{i+1} are connected 8-directionally (ie., they are different and share an edge or corner)
  • C_1 is at location (0, 0) (ie. has value grid[0][0])
  • C_k is at location (N-1, N-1) (ie. has value grid[N-1][N-1])
  • If C_i is located at (r, c), then grid[r][c] is empty (ie. grid[r][c] == 0).

Return the length of the shortest such clear path from top-left to bottom-right.  If such a path does not exist, return -1.

 

Example 1:

Input: [[0,1],[1,0]]


Output: 2

Example 2:

Input: [[0,0,0],[1,1,0],[1,1,0]]


Output: 4

 

Note:

  1. 1 <= grid.length == grid[0].length <= 100
  2. grid[r][c] is 0 or 1

1091. Shortest Path in Binary Matrix
struct Solution;
use std::collections::VecDeque;

type Point = (usize, usize);

impl Solution {
    fn shortest_path_binary_matrix(grid: Vec<Vec<i32>>) -> i32 {
        let n = grid.len();
        let m = grid[0].len();
        let mut queue: VecDeque<Point> = VecDeque::new();
        if grid[0][0] == 1 || grid[(n - 1) as usize][(m - 1) as usize] == 1 {
            return -1;
        }
        let mut visited: Vec<Vec<bool>> = vec![vec![false; m]; n];
        visited[0][0] = true;
        queue.push_back((0, 0));
        let mut distance = 0;
        let offsets = [
            (-1, -1),
            (0, -1),
            (1, -1),
            (-1, 1),
            (0, 1),
            (1, 1),
            (-1, 0),
            (1, 0),
        ];
        while !queue.is_empty() {
            distance += 1;
            let queue_size = queue.len();
            for _ in 0..queue_size {
                let p = queue.pop_front().unwrap();
                if p.0 == n - 1 && p.1 == m - 1 {
                    return distance;
                } else {
                    for offset in &offsets {
                        let x = p.0 as i32 + offset.0;
                        let y = p.1 as i32 + offset.1;
                        if !(x < 0 || x >= n as i32 || y < 0 || y >= m as i32)
                            && grid[x as usize][y as usize] == 0
                            && !visited[x as usize][y as usize]
                        {
                            visited[x as usize][y as usize] = true;
                            queue.push_back((x as usize, y as usize));
                        }
                    }
                }
            }
        }
        -1
    }
}

#[test]
fn test() {
    let grid = vec_vec_i32![[0, 1], [1, 0]];
    let res = 2;
    assert_eq!(Solution::shortest_path_binary_matrix(grid), res);
    let grid = vec_vec_i32![[0, 0, 0], [1, 1, 0], [1, 1, 0]];
    let res = 4;
    assert_eq!(Solution::shortest_path_binary_matrix(grid), res);
    let grid = vec_vec_i32![[1, 0, 0], [1, 1, 0], [1, 1, 0]];
    let res = -1;
    assert_eq!(Solution::shortest_path_binary_matrix(grid), res);
    let grid = vec_vec_i32![
        [0, 1, 0, 0, 0, 0],
        [0, 1, 1, 1, 1, 1],
        [0, 0, 0, 0, 1, 1],
        [0, 1, 0, 0, 0, 1],
        [1, 0, 0, 1, 0, 1],
        [0, 0, 1, 0, 1, 0]
    ];
    let res = 7;
    assert_eq!(Solution::shortest_path_binary_matrix(grid), res);
}