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Formatted question description: https://leetcode.ca/all/1926.html

1926. Nearest Exit from Entrance in Maze

Level

Medium

Description

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 = [entrance_row, entrance_col] 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:

Image text

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:

Image text

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.

Example 3:

Image text

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 <= entrance_row < m
  • 0 <= entrance_col < n
  • entrance will always be an empty cell.

Solution

Do breadth-first search in maze starting from entrance. The cell entrance is visited first and does not count as an exit. Once the first empty cell that is on the border is visited, return the number of steps.

  • class Solution {
    public int nearestExit(char[][] maze, int[] entrance) {
        int[][] directions = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
        int m = maze.length, n = maze[0].length;
        boolean[][] visited = new boolean[m][n];
        visited[entrance[0]][entrance[1]] = true;
        Queue<int[]> queue = new LinkedList<int[]>();
        queue.offer(entrance);
        int steps = 0;
        while (!queue.isEmpty()) {
            steps++;
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                int[] cell = queue.poll();
                int row = cell[0], column = cell[1];
                for (int[] direction : directions) {
                    int newRow = row + direction[0], newColumn = column + direction[1];
                    if (newRow >= 0 && newRow < m && newColumn >= 0 && newColumn < n && !visited[newRow][newColumn] && maze[newRow][newColumn] == '.') {
                        if (newRow == 0 || newRow == m - 1 || newColumn == 0 || newColumn == n - 1)
                            return steps;
                        visited[newRow][newColumn] = true;
                        queue.offer(new int[]{newRow, newColumn});
                    }
                }
            }
        }
        return -1;
    }
}

############

class Solution {
    public int nearestExit(char[][] maze, int[] entrance) {
        int m = maze.length;
        int n = maze[0].length;
        Deque<int[]> q = new ArrayDeque<>();
        q.offer(entrance);
        maze[entrance[0]][entrance[1]] = '+';
        int ans = 0;
        int[] dirs = {-1, 0, 1, 0, -1};
        while (!q.isEmpty()) {
            ++ans;
            for (int k = q.size(); k > 0; --k) {
                int[] p = q.poll();
                int i = p[0], j = p[1];
                for (int l = 0; l < 4; ++l) {
                    int x = i + dirs[l], y = j + dirs[l + 1];
                    if (x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == '.') {
                        if (x == 0 || x == m - 1 || y == 0 || y == n - 1) {
                            return ans;
                        }
                        q.offer(new int[] {x, y});
                        maze[x][y] = '+';
                    }
                }
            }
        }
        return -1;
    }
}

  • // OJ: https://leetcode.com/problems/nearest-exit-from-entrance-in-maze/
// Time: O(MN)
// Space: O(min(M, N)^2)
class Solution {
public:
    int nearestExit(vector<vector<char>>& A, vector<int>& E) {
        int dirs[4][2] = { {0,1},{0,-1},{-1,0},{1,0} }, step = 0, M = A.size(), N = A[0].size();
        queue<pair<int, int>> q;
        A[E[0]][E[1]] = '+';
        q.emplace(E[0], E[1]);
        while (q.size()) {
            int cnt = q.size();
            while (cnt--) {
                auto [x, y] = q.front();
                q.pop();
                if ((x != E[0] || y != E[1]) && (x == 0 || x == M - 1 || y == 0 || y == N - 1)) return step;
                for (auto &[dx, dy] : dirs) {
                    int a = x + dx, b = y + dy;
                    if (a < 0 || b < 0 || a >= M || b >= N || A[a][b] == '+') continue;
                    A[a][b] = '+';
                    q.emplace(a, b);
                }
            }
            ++step;
        }
        return -1;
    }
};

  • class Solution:
    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
        m, n = len(maze), len(maze[0])
        i, j = entrance
        q = deque([(i, j)])
        maze[i][j] = '+'
        ans = 0
        while q:
            ans += 1
            for _ in range(len(q)):
                i, j = q.popleft()
                for a, b in [[0, -1], [0, 1], [-1, 0], [1, 0]]:
                    x, y = i + a, j + b
                    if 0 <= x < m and 0 <= y < n and maze[x][y] == '.':
                        if x == 0 or x == m - 1 or y == 0 or y == n - 1:
                            return ans
                        q.append((x, y))
                        maze[x][y] = '+'
        return -1

############

# 1926. Nearest Exit from Entrance in Maze
# https://leetcode.com/problems/nearest-exit-from-entrance-in-maze

class Solution:
    def nearestExit(self, maze: List[List[str]], entrance: List[int]) -> int:
        rows, cols = len(maze), len(maze[0])
        ox, oy = entrance
        if maze[ox][oy] == '+': return -1
        
        queue = collections.deque([[ox, oy, 0]])
        visited = set([(ox, oy)])
        
        while queue:
            x, y, steps = queue.popleft()
            if steps > 0 and (x == 0 or y == 0 or x == rows - 1 or y == cols - 1): return steps
            
            for dx, dy in ((x + 1, y), (x - 1, y), (x, y + 1), (x, y - 1)):
                if 0 <= dx < rows and 0 <= dy < cols and (dx, dy) not in visited and maze[dx][dy] == '.':
                    visited.add((dx, dy))
                    queue.append((dx, dy, steps + 1))
        
        return -1


  • func nearestExit(maze [][]byte, entrance []int) int {
	m, n := len(maze), len(maze[0])
	q := [][]int{entrance}
	maze[entrance[0]][entrance[1]] = '+'
	ans := 0
	dirs := []int{-1, 0, 1, 0, -1}
	for len(q) > 0 {
		ans++
		for k := len(q); k > 0; k-- {
			p := q[0]
			q = q[1:]
			for l := 0; l < 4; l++ {
				x, y := p[0]+dirs[l], p[1]+dirs[l+1]
				if x >= 0 && x < m && y >= 0 && y < n && maze[x][y] == '.' {
					if x == 0 || x == m-1 || y == 0 || y == n-1 {
						return ans
					}
					q = append(q, []int{x, y})
					maze[x][y] = '+'
				}
			}
		}
	}
	return -1
}

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