<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>>& arr, vector<int>& entrance) {
        int m=arr.size();
        int n=arr[0].size();
        vector<vector<int> > dp(m,vector<int>(n,-1));
        dp[entrance[0]][entrance[1]]=0;  //src;
        vector<int> dx={1,-1,0,0};
        vector<int> dy={0,0,1,-1};

        queue<pair<int,int> > q;
        q.push({entrance[0],entrance[1]});

        while(!q.empty()){
            int x=q.front().first;
            int y=q.front().second;
            q.pop();

            for(int ii=0;ii<4;ii++){
                int fx=x+dx[ii];
                int fy=y+dy[ii];
                if(fx>=0 && fx<m && fy>=0 && fy<n && arr[fx][fy]=='.' && dp[fx][fy]==-1){
                    dp[fx][fy]=dp[x][y]+1;
                    q.push({fx,fy});

                }
                
            }
        }   
            vector<int> temp; 
            int xx=-1,yy=-1,val=INT_MAX;
            for(int i=0;i<m;i++){
                // i,0    &&    i.n-1
                temp={i,0};
                if(entrance!=temp && dp[temp[0]][temp[1]]!=-1){
                    val=min(val,dp[i][0]);
                    xx=i;yy=0;
                }
            }

            for(int i=0;i<m;i++){
                // i,0    &&    i.n-1

                temp={i,n-1};
                if(entrance!=temp && dp[temp[0]][temp[1]]!=-1){
                    val=min(val,dp[i][n-1]);
                    xx=i;yy=n-1;
                }
            }

            //0,j   &&    m-1,j
            for(int j=0;j<n;j++){
                temp={0,j};
                if(entrance!=temp && dp[temp[0]][temp[1]]!=-1){
                    val=min(val,dp[0][j]);
                    xx=0;yy=j;
                }
            }//0,j   &&    m-1,j
            for(int j=0;j<n;j++){
                temp={m-1,j};
                if(entrance!=temp && dp[temp[0]][temp[1]]!=-1){
                    val=min(val,dp[m-1][j]);
                    xx=m-1;yy=j;
                }
            }

        if(xx==-1) return -1;
        return val;
        
        
    }
};
39721