Nearest Exit from Entrance in Maze

Problem

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.

Source: Leetcode 1926. Nearest Exit from Entrance in Maze

Solution

This is a trivial BFS problem. If you do not know, BFS stands for Breadth First Search. This graph algorithm explores neighbouring nodes as one layer at a time. Then for each node of the first layer, it explores the second layer. This way, it doesn’t go deep into one path like Depth First Search algorithm does.

To apply BFS, we will use a queue and push the source coordinates into queue. Until queue is empty or any result node is not found, we will continue exploring layer by layer and keep track of visited nodes. Also we will update the distance of a newly explored node with respect to its parent node. We will do like, distance of source = distance of parent + 1.

Here is a simple C++ code for your understanding.

class Solution {
public:
    int nearestExit(vector<vector<char>>& maze, vector<int>& entrance) {
        queue<pair<int, int>> q;
        q.push({entrance[0], entrance[1]}); // push the source node to queue
        
        int R = maze.size();
        int C = maze[0].size();
        
        vector<vector<int>> dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        vector<vector<int>> visited(101, vector<int>(101, -1)); // to keep track of explored or unexplored nodes, unexplored = -1
        visited[entrance[0]][entrance[1]] = 0;
        
        while(!q.empty()){
            pair<int, int> cur = q.front(); // current parent node
            q.pop();
            for(int i = 0; i < 4; i++){
                int new_x = cur.first + dir[i][0]; // Coordinates of New child to explored 
                int new_y = cur.second + dir[i][1];
                
                if(0 <= new_x && new_x < R && 0 <= new_y && new_y < C){
                    if(maze[new_x][new_y] == '.'){ // if it is a valid cell to move
                        if(visited[new_x][new_y] == -1){ // and not visited already
                            visited[new_x][new_y] = visited[cur.first][cur.second] + 1; // update child distance with +1 of parent (cur)
                            if(new_x == 0 || new_x == R - 1 || new_y == 0 || new_y == C - 1) // Terminal condition found!!! Let's return :)
                                return visited[new_x][new_y];
                            q.push({new_x, new_y}); //push this child to the queue
                        }
                    }
                }
            }
        }
        return -1; // cannot go our of maze, return -1!
    }
};

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