<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>

int sf[401][401];
int dp[401][401];

class Solution {
public:
    void safecompute(vector<vector<int>>& grid) {
         int n = grid.size();
         queue<pair<int,int>> q;
         for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                if(grid[i][j] == 1) {
                    q.push({i,j});
                }
            }
         }
         int cnt = 0;
         vector<vector<int>> visited(n,vector<int>(n));
         while(!q.empty()) {
            int size = q.size();
            while(size--) {
                int x = q.front().first;
                int y = q.front().second;
                q.pop();
                if(visited[x][y] == 1) continue;
                sf[x][y] = cnt;

                vector<int> vx = {0,0,-1,1};
                vector<int> vy = {1,-1,0,0};
                visited[x][y] = 1;
                for(int i=0;i<4;i++) {
                    int nx = x+vx[i];
                    int ny = y+vy[i];

                    if(nx >= 0  &&  ny >=0   &&  nx < n  &&  ny < n  &&  visited[nx][ny] == 0) {
                        q.push({nx, ny});
                    }
                }
            }
            cnt++;
         }
    }
    
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();

        for(int i=0;i<n;i++) {
            for(int j=0;j<n;j++) {
                sf[i][j] = 1e8;
            }
        }
        
        safecompute(grid);
        priority_queue<pair<int,pair<int,int>>> pq;
        pq.push({sf[0][0], {0,0}});
        vector<vector<int>> visited(n, vector<int>(n));
        while(!pq.empty()) {
            int safeness = pq.top().first;
            int x = pq.top().second.first;
            int y = pq.top().second.second;

            pq.pop();
            if(x == n-1  &&  y == n-1) {
                return safeness;
            }
            if(visited[x][y] == 1) continue;
            visited[x][y] = 1;
            vector<int> vx = {0,0,-1,1};
            vector<int> vy = {1,-1,0,0};

            for(int k=0;k<4;k++) {
                int nx = x+vx[k];
                int ny = y+vy[k];

                if(nx >=0  &&  nx < n  &&  ny >=0  &&  ny < n  &&  visited[nx][ny] == 0) {
                    pq.push({min(safeness, sf[nx][ny]), {nx,ny}});
                }
            }
        }

        return 0;
    }
};
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