You are given an m x n
matrix maze
(0-indexed) with empty cells (represented as '.'
) and walls (represented as '+'
). You are also given the entrance
of the maze, where entrance = [entrancerow, entrancecol]
denotes the row and column of the cell you are initially standing at.
In one step, you can move one cell up, down, left, or right. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the nearest exit from the entrance
. An exit is defined as an empty cell that is at the border of the maze
. The entrance
does not count as an exit.
Return the number of steps in the shortest path from the entrance
to the nearest exit, or -1
if no such path exists.
Input: maze = [["+","+",".","+"],[".",".",".","+"],["+","+","+","."]], entrance = [1,2] Output: 1 Explanation: There are 3 exits in this maze at [1,0], [0,2], and [2,3]. Initially, you are at the entrance cell [1,2]. - You can reach [1,0] by moving 2 steps left. - You can reach [0,2] by moving 1 step up. It is impossible to reach [2,3] from the entrance. Thus, the nearest exit is [0,2], which is 1 step away.
Input: maze = [["+","+","+"],[".",".","."],["+","+","+"]], entrance = [1,0] Output: 2 Explanation: There is 1 exit in this maze at [1,2]. [1,0] does not count as an exit since it is the entrance cell. Initially, you are at the entrance cell [1,0]. - You can reach [1,2] by moving 2 steps right. Thus, the nearest exit is [1,2], which is 2 steps away.
Input: maze = [[".","+"]], entrance = [0,0] Output: -1 Explanation: There are no exits in this maze.
maze.length == m
maze[i].length == n
1 <= m, n <= 100
maze[i][j]
is either'.'
or'+'
.entrance.length == 2
0 <= entrancerow < m
0 <= entrancecol < n
entrance
will always be an empty cell.
use std::collections::VecDeque;
impl Solution {
pub fn nearest_exit(maze: Vec<Vec<char>>, entrance: Vec<i32>) -> i32 {
let m = maze.len() as i32;
let n = maze[0].len() as i32;
let mut maze = maze;
let mut cells = VecDeque::from([(entrance[0], entrance[1], 0)]);
maze[cells[0].0 as usize][cells[0].1 as usize] = '+';
while let Some((row, col, steps)) = cells.pop_front() {
if (row != entrance[0] || col != entrance[1])
&& (row == 0 || col == 0 || row == m - 1 || col == n - 1)
{
return steps;
}
for (r, c) in [(-1, 0), (1, 0), (0, -1), (0, 1)] {
let row = row + r;
let col = col + c;
if row >= 0
&& row < m
&& col >= 0
&& col < n
&& maze[row as usize][col as usize] == '.'
{
cells.push_back((row, col, steps + 1));
maze[row as usize][col as usize] = '+';
}
}
}
-1
}
}