<p>You are given a <strong>0-indexed</strong> 2D matrix <code>grid</code> of size <code>n x n</code>, where <code>(r, c)</code> represents:</p>

<ul>
	<li>A cell containing a thief if <code>grid[r][c] = 1</code></li>
	<li>An empty cell if <code>grid[r][c] = 0</code></li>
</ul>

<p>You are initially positioned at cell <code>(0, 0)</code>. In one move, you can move to any adjacent cell in the grid, including cells containing thieves.</p>

<p>The <strong>safeness factor</strong> of a path on the grid is defined as the <strong>minimum</strong> manhattan distance from any cell in the path to any thief in the grid.</p>

<p>Return <em>the <strong>maximum safeness factor</strong> of all paths leading to cell </em><code>(n - 1, n - 1)</code><em>.</em></p>

<p>An <strong>adjacent</strong> cell of cell <code>(r, c)</code>, is one of the cells <code>(r, c + 1)</code>, <code>(r, c - 1)</code>, <code>(r + 1, c)</code> and <code>(r - 1, c)</code> if it exists.</p>

<p>The <strong>Manhattan distance</strong> between two cells <code>(a, b)</code> and <code>(x, y)</code> is equal to <code>|a - x| + |b - y|</code>, where <code>|val|</code> denotes the absolute value of val.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example1.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,1]]
<strong>Output:</strong> 0
<strong>Explanation:</strong> All paths from (0, 0) to (n - 1, n - 1) go through the thieves in cells (0, 0) and (n - 1, n - 1).
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example2.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,1],[0,0,0],[0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 2) is cell (0, 0). The distance between them is | 0 - 0 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2023/07/02/example3.png" style="width: 362px; height: 242px;" />
<pre>
<strong>Input:</strong> grid = [[0,0,0,1],[0,0,0,0],[0,0,0,0],[1,0,0,0]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The path depicted in the picture above has a safeness factor of 2 since:
- The closest cell of the path to the thief at cell (0, 3) is cell (1, 2). The distance between them is | 0 - 1 | + | 3 - 2 | = 2.
- The closest cell of the path to the thief at cell (3, 0) is cell (3, 2). The distance between them is | 3 - 3 | + | 0 - 2 | = 2.
It can be shown that there are no other paths with a higher safeness factor.
</pre>

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

<ul>
	<li><code>1 &lt;= grid.length == n &lt;= 400</code></li>
	<li><code>grid[i].length == n</code></li>
	<li><code>grid[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
	<li>There is at least one thief in the <code>grid</code>.</li>
</ul>

class Solution {
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
           queue<pair<int,int>> q;
           int n=grid.size();
           for(int i=0;i<n;i++){
             for(int j=0;j<n;j++){
                  if(grid[i][j]==1){
                    q.push({i,j});
                    grid[i][j]=0;
                  }else{
                    grid[i][j]=500;
                  }
             }
           }
           int dr[]={-1,0,1,0};
           int dc[]={0,-1,0,1};
           while(!q.empty()){
               int x=q.front().first;
               int y=q.front().second;
               q.pop();
               for(int i=0;i<4;i++){
                   int r=x+dr[i];
                   int c=y+dc[i];
                   if(r>=0 && c>=0 && r<n && c<n && grid[r][c]>grid[x][y]+1){
                        grid[r][c]=1+grid[x][y];
                        q.push({r,c});
                   }
               }
           }
           vector<vector<int>> dist(n, vector<int> (n,0));
           dist[0][0]=grid[0][0];
           priority_queue<vector<int>> pq;
           pq.push({dist[0][0],0,0});
        //    int dr[]={-1,0,1,0};
        //    int dc[]={0,-1,0,1};
           while(!pq.empty()){
               int sf=pq.top()[0];
               int x=pq.top()[1];
               int y=pq.top()[2];
               pq.pop();
               for(int i=0;i<4;i++){
                   int r=x+dr[i];
                   int c=y+dc[i];
                   if(r>=0 && c>=0 && r<n && c<n && min(grid[r][c],sf)>dist[r][c]){
                        dist[r][c]=min(grid[r][c],sf);
                        pq.push({dist[r][c],r,c});
                   }
               }
           }
           return dist[n-1][n-1];
    }
};
194408