Medium
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.
Example 1:
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.
Example 2:
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].
Thus, the nearest exit is [1,2], which is 2 steps away.
Example 3:
Input: maze = [[”.”,”+”]], entrance = [0,0]
Output: -1
Explanation: There are no exits in this maze.
Constraints:
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.import java.util.LinkedList;
import java.util.Queue;
public class Solution {
public int nearestExit(char[][] maze, int[] entrance) {
int m = maze.length;
int n = maze[0].length;
int[] directions = new int[] {0, 1, 0, -1, 0};
Queue<int[]> queue = new LinkedList<>();
queue.offer(new int[] {entrance[0], entrance[1], 0});
boolean[][] visited = new boolean[m][n];
visited[entrance[0]][entrance[1]] = true;
int shortestSteps = m * n;
while (!queue.isEmpty()) {
int[] curr = queue.poll();
for (int i = 0; i < directions.length - 1; i++) {
int nextX = curr[0] + directions[i];
int nextY = curr[1] + directions[i + 1];
if (nextX >= 0
&& nextX < m
&& nextY >= 0
&& nextY < n
&& maze[nextX][nextY] == '.'
&& !visited[nextX][nextY]) {
visited[nextX][nextY] = true;
if (nextX == 0 || nextX == m - 1 || nextY == 0 || nextY == n - 1) {
shortestSteps = Math.min(shortestSteps, curr[2] + 1);
} else {
queue.offer(new int[] {nextX, nextY, curr[2] + 1});
}
}
}
}
return shortestSteps == m * n ? -1 : shortestSteps;
}
}