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Biweekly Contest 56 Q2
Breadth-First Search
Array
Matrix

1926. Nearest Exit from Entrance in Maze

中文文档

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 = [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].
- You can reach [1,2] by moving 2 steps right.
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.

Solutions

Solution 1: BFS

We can start from the entrance and perform a breadth-first search (BFS). Each time we reach a new empty cell, we mark it as visited and add it to the queue until we find an empty cell on the boundary, then return the number of steps.

Specifically, we define a queue $q$, initially adding $\textit{entrance}$ to the queue. We define a variable $\textit{ans}$ to record the number of steps, initially set to $1$. Then we start the BFS. In each round, we take out all elements from the queue and traverse them. For each element, we try to move in four directions. If the new position is an empty cell, we add it to the queue and mark it as visited. If the new position is an empty cell on the boundary, we return $\textit{ans}$. If the queue is empty, we return $-1$. After this round of search, we increment $\textit{ans}$ by one and continue to the next round of search.

If we finish the traversal without finding an empty cell on the boundary, we return $-1$.

The time complexity is $O(m \times n)$, and the space complexity is $O(m \times n)$. Here, $m$ and $n$ are the number of rows and columns in the maze, respectively.

Python3

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

Java

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

C++

class Solution {
public:
    int nearestExit(vector<vector<char>>& maze, vector<int>& entrance) {
        int m = maze.size(), n = maze[0].size();
        int dirs[5] = {-1, 0, 1, 0, -1};
        queue<pair<int, int>> q;
        q.emplace(entrance[0], entrance[1]);
        maze[entrance[0]][entrance[1]] = '+';
        for (int ans = 1; !q.empty(); ++ans) {
            for (int k = q.size(); k; --k) {
                auto [i, j] = q.front();
                q.pop();
                for (int d = 0; d < 4; ++d) {
                    int x = i + dirs[d], y = j + dirs[d + 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;
                        }
                        maze[x][y] = '+';
                        q.emplace(x, y);
                    }
                }
            }
        }
        return -1;
    }
};

Go

func nearestExit(maze [][]byte, entrance []int) int {
	m, n := len(maze), len(maze[0])
	q := [][2]int{{entrance[0], entrance[1]}}
	maze[entrance[0]][entrance[1]] = '+'
	dirs := []int{-1, 0, 1, 0, -1}
	for ans := 1; 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, [2]int{x, y})
					maze[x][y] = '+'
				}
			}
		}
	}
	return -1
}

TypeScript

function nearestExit(maze: string[][], entrance: number[]): number {
    const dir = [0, 1, 0, -1, 0];
    const q = [[...entrance, 0]];
    maze[entrance[0]][entrance[1]] = '+';
    for (const [i, j, ans] of q) {
        for (let d = 0; d < 4; d++) {
            const [x, y] = [i + dir[d], j + dir[d + 1]];
            const v = maze[x]?.[y];
            if (!v && ans) {
                return ans;
            }
            if (v === '.') {
                q.push([x, y, ans + 1]);
                maze[x][y] = '+';
            }
        }
    }
    return -1;
}