<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:
    vector<int>delr={-1 ,0 ,1 ,0};
    vector<int>delc={0 ,-1 ,0 ,1};
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int r=grid.size();
        int c=grid[0].size();
        vector<vector<int>>dis(r ,vector<int>(c ,1e9));
        vector<vector<bool>>vis(r ,vector<bool>(c ,false));
        queue<pair<int,pair<int ,int>>>q;
        for(int i=0;i<r;i++){
            for(int j=0;j<c;j++){
                if(grid[i][j]==1 &&!vis[i][j] ){
                    vis[i][j]=true;
                    dis[i][j]=0;
                    q.push({0 ,{i ,j}});
                }
            }
        }
        while(!q.empty()){
            auto it=q.front();
            q.pop();
            int cur_dis= it.first;
            int row =it.second.first;
            int col=it.second.second;
            dis[row][col]=cur_dis;
            for(int i=0;i<4;i++){
                int nrow = row+delr[i];
                int ncol = col+delc[i];
                if(nrow>=0 && ncol>=0 && nrow<r && ncol<c && !vis[nrow][ncol]  ){
                    vis[nrow][ncol]=true;
                    q.push({cur_dis+1 ,{nrow , ncol}});
                }
            }
        }



        priority_queue< pair<int,pair<int,int>> >maxHeap;
        vector<vector<bool>>vis2(r ,vector<bool>(c ,false));
        vis[0][0]=true;
        int ans=0;
        maxHeap.push({dis[0][0],{0,0}});
 
        while(!maxHeap.empty()){
            auto it=maxHeap.top();
            maxHeap.pop();
            int d = it.first;
            int row=it.second.first;
            int col =it.second.second;
            if(row==r-1 && col==c-1){
                return d;
            }
            for(int k=0;k<4;k++){
                int nrow = row+delr[k];
                int ncol = col+delc[k];
                if( nrow>=0 && ncol >=0 && nrow<r && ncol <c && !vis2[nrow][ncol] ){
                    maxHeap.push({ min(d ,dis[nrow][ncol]) ,{nrow ,ncol}});
                    vis2[nrow][ncol]=true;
                }
            }
        }

        return ans;

    }
};
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