<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();
        int n = maze[0].size();


        queue<pair<int, int>> q;

        vector<vector<int>> distance(m, vector<int>(n, INT_MAX));
        vector<vector<bool>> visited(m, vector<bool>(n, false));

        q.push({entrance[0], entrance[1]});
        visited[entrance[0]][entrance[1]]=true;
        distance[entrance[0]][entrance[1]]=0;

        while(!q.empty())
        {
            pair<int, int> p = q.front();
            int r=p.first;
            int c=p.second;

            if(c-1>=0 && maze[r][c-1]=='.' && !visited[r][c-1])
            {
                distance[r][c-1]=min(distance[r][c-1], distance[r][c]+1);
                q.push({r,c-1});
                visited[r][c-1]=true;
            }
            if(c+1<n && maze[r][c+1]=='.' && !visited[r][c+1])
            {
                distance[r][c+1]=min(distance[r][c+1], distance[r][c]+1);
                q.push({r,c+1});
                visited[r][c+1]=true;
            }

            if(r-1>=0 && maze[r-1][c]=='.' && !visited[r-1][c])
            {
                distance[r-1][c]=min(distance[r-1][c], distance[r][c]+1);
                q.push({r-1,c});
                visited[r-1][c]=true;
            }
            if(r+1<m && maze[r+1][c]=='.' && !visited[r+1][c])
            {
                distance[r+1][c]=min(distance[r+1][c], distance[r][c]+1);
                q.push({r+1,c});
                visited[r+1][c]=true;
            }



            q.pop();
        }

        int ans = INT_MAX;
        for(int i=0;i<m;i++)
        {
            if(distance[i][0]!=0)
                ans = min(distance[i][0], ans);
        }
        for(int i=0;i<m;i++)
        {
            if(distance[i][n-1]!=0)
                ans = min(distance[i][n-1], ans);
        }
        for(int j=0;j<n;j++)
        {
            if(distance[0][j]!=0)
                ans = min(distance[0][j], ans);
        }
        for(int j=0;j<n;j++)
        {
            if(distance[m-1][j]!=0)
                ans = min(distance[m-1][j], ans);
        }



        
        if(ans == INT_MAX)
            return -1;
        return ans;
    }
};
40165