<p>You are given an <code>m x n</code> matrix <code>maze</code> (<strong>0-indexed</strong>) with empty cells (represented as <code>&#39;.&#39;</code>) and walls (represented as <code>&#39;+&#39;</code>). You are also given the <code>entrance</code> of the maze, where <code>entrance = [entrance<sub>row</sub>, entrance<sub>col</sub>]</code> denotes the row and column of the cell you are initially standing at.</p>

<p>In one step, you can move one cell <strong>up</strong>, <strong>down</strong>, <strong>left</strong>, or <strong>right</strong>. You cannot step into a cell with a wall, and you cannot step outside the maze. Your goal is to find the <strong>nearest exit</strong> from the <code>entrance</code>. An <strong>exit</strong> is defined as an <strong>empty cell</strong> that is at the <strong>border</strong> of the <code>maze</code>. The <code>entrance</code> <strong>does not count</strong> as an exit.</p>

<p>Return <em>the <strong>number of steps</strong> in the shortest path from the </em><code>entrance</code><em> to the nearest exit, or </em><code>-1</code><em> if no such path exists</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/nearest1-grid.jpg" style="width: 333px; height: 253px;" />
<pre>
<strong>Input:</strong> maze = [[&quot;+&quot;,&quot;+&quot;,&quot;.&quot;,&quot;+&quot;],[&quot;.&quot;,&quot;.&quot;,&quot;.&quot;,&quot;+&quot;],[&quot;+&quot;,&quot;+&quot;,&quot;+&quot;,&quot;.&quot;]], entrance = [1,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/nearesr2-grid.jpg" style="width: 253px; height: 253px;" />
<pre>
<strong>Input:</strong> maze = [[&quot;+&quot;,&quot;+&quot;,&quot;+&quot;],[&quot;.&quot;,&quot;.&quot;,&quot;.&quot;],[&quot;+&quot;,&quot;+&quot;,&quot;+&quot;]], entrance = [1,0]
<strong>Output:</strong> 2
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/06/04/nearest3-grid.jpg" style="width: 173px; height: 93px;" />
<pre>
<strong>Input:</strong> maze = [[&quot;.&quot;,&quot;+&quot;]], entrance = [0,0]
<strong>Output:</strong> -1
<strong>Explanation:</strong> There are no exits in this maze.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>maze.length == m</code></li>
	<li><code>maze[i].length == n</code></li>
	<li><code>1 &lt;= m, n &lt;= 100</code></li>
	<li><code>maze[i][j]</code> is either <code>&#39;.&#39;</code> or <code>&#39;+&#39;</code>.</li>
	<li><code>entrance.length == 2</code></li>
	<li><code>0 &lt;= entrance<sub>row</sub> &lt; m</code></li>
	<li><code>0 &lt;= entrance<sub>col</sub> &lt; n</code></li>
	<li><code>entrance</code> will always be an empty cell.</li>
</ul>

class Solution {
public:
    int nearestExit(vector<vector<char>>& maze, vector<int>& entrance) {
        int m = maze.size(); // number of rows
        int n = maze[0].size(); // number of columns
        
        // BFS queue: stores row, column, and steps taken so far
        queue<pair<int, int>> q;
        q.push({entrance[0], entrance[1]});
        
        // Visited array to avoid revisiting cells
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        visited[entrance[0]][entrance[1]] = true;
        
        // Direction vectors for moving up, down, left, and right
        vector<pair<int, int>> directions = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};
        
        int steps = 0;
        
        // Perform BFS
        while (!q.empty()) {
            int qSize = q.size();
            steps++; // Increment step at each BFS level

            // Process all nodes at the current level
            for (int i = 0; i < qSize; i++) {
                auto [currRow, currCol] = q.front();
                q.pop();
                
                // Try all 4 directions
                for (auto dir : directions) {
                    int newRow = currRow + dir.first;
                    int newCol = currCol + dir.second;

                    if (isValid(maze, visited, newRow, newCol, m, n)) {
                        // If it's on the border and not the entrance, return steps
                        if (isExit(newRow, newCol, m, n) && !(newRow == entrance[0] && newCol == entrance[1])) {
                            return steps;
                        }
                        
                        // Mark as visited and add to queue
                        visited[newRow][newCol] = true;
                        q.push({newRow, newCol});
                    }
                }
            }
        }
        
        return -1; // No exit found
    }

    bool isValid(vector<vector<char>>& maze, vector<vector<bool>>& visited, int r, int c, int m, int n) {
        // Check if the position is within bounds and is an empty cell, not visited yet
        return r >= 0 && r < m && c >= 0 && c < n && maze[r][c] == '.' && !visited[r][c];
    }
    
    bool isExit(int r, int c, int m, int n) {
        // Check if the cell is on the border of the maze
        return r == 0 || r == m - 1 || c == 0 || c == n - 1;
    }
};

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