<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 {
    vector<int>dir={0,1,0,-1,0};
    bool isValid(int row,int col,int n){
        return row>=0 && col>=0 && row<n && col<n;
    }
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = (int)grid.size();
        if(n==0)return -1;
        vector<vector<int>>safeness(n,vector<int>(n,INT_MAX));
        queue<vector<int>>q;
        for (int row = 0; row < n; ++row) {
            for (int col = 0; col < n; ++col) {
                if(grid[row][col]==1){
                    safeness[row][col]=0;
                    q.push({row,col,0});
                }
            }
        }
        while(!q.empty()){
            auto cell = q.front();
            q.pop();
            auto row = cell[0], col = cell[1] ,dis = cell[2];
            for (int i = 0; i < 4; ++i) {
                int nextRow = row+dir[i];
                int nextCol = col+dir[i+1];
                if(!isValid(nextRow,nextCol,n) || safeness[nextRow][nextCol]!=INT_MAX)continue;
                safeness[nextRow][nextCol] = dis+1;
                q.push({nextRow,nextCol,safeness[nextRow][nextCol]});
            }
        }
        vector<vector<int>>vis(n,vector<int>(n));
        deque<vector<int>>d;
        int minSafety = safeness[0][0];
        d.push_front({safeness[0][0],0,0});
        vis[0][0]=1;
        while(!d.empty()){
            auto cell = d.front();
            d.pop_front();
            int safety=cell[0], row = cell[1], col=cell[2];
            minSafety = min(minSafety,safety);
            if(row==n-1 && col==n-1)break;
            for (int i = 0; i < 4; ++i) {
                int nextRow = row+dir[i];
                int nextCol = col+dir[i+1];
                if(!isValid(nextRow,nextCol,n) || vis[nextRow][nextCol])continue;
                vis[nextRow][nextCol]=1;
                if(safeness[nextRow][nextCol]>=minSafety){
                    d.push_front({safeness[nextRow][nextCol],nextRow,nextCol});
                }
                else{
                    d.push_back({safeness[nextRow][nextCol],nextRow,nextCol});
                }
            }
        }
        return minSafety;
    }
};

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