<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 findManhattan(vector<vector<int>> &grid, int pr, int pc) {
        int n = grid.size();
        int ans = 3*n;
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<n; j++) {
                if (grid[i][j] == 1) ans = min(ans, abs(i-pr)+abs(j-pc));
            }
        }
        return ans;
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> md(n, vector<int>(n, 3*n));
        md[0][0] = findManhattan(grid, 0, 0);
        for(int i = 0; i<n; i++) {
            for(int j = 0; j<n; j++) {
                int tmp = 3*n;
                if (grid[i][j] == 1) tmp = 0;
                if (i>0) tmp = min(tmp, md[i-1][j]+1);
                if (j>0) tmp = min(tmp, md[i][j-1]+1);
                md[i][j] = min(md[i][j], tmp);
            }
        }
        md[n-1][n-1] = findManhattan(grid, n-1, n-1);
        for(int i = n-1; i>=0; i--) {
            for(int j = n-1; j>=0; j--) {
                int tmp = 3*n;
                if (grid[i][j] == 1) tmp = 0;
                if (i<n-1) tmp = min(tmp, md[i+1][j]+1);
                if (j<n-1) tmp = min(tmp, md[i][j+1]+1);
                md[i][j] = min(md[i][j], tmp);
            }
        }
        vector<vector<int>> ans(n, vector<int>(n, -1));
        priority_queue<pair<int, pair<int,int>>> pq;
        pq.push({md[0][0], {0,0}});
        while(!pq.empty()) {
            pair<int, pair<int,int>> p = pq.top();
            pq.pop();
            int sf = p.first;
            int pr = p.second.first;
            int pc = p.second.second;
            if (sf <= ans[pr][pc]) continue;
            ans[pr][pc] = sf;
            if (pr > 0) pq.push({min(sf, md[pr-1][pc]),{pr-1, pc}});
            if (pc > 0) pq.push({min(sf, md[pr][pc-1]),{pr, pc-1}});
            if (pr < n-1) pq.push({min(sf, md[pr+1][pc]),{pr+1, pc}});
            if (pc < n-1) pq.push({min(sf, md[pr][pc+1]),{pr, pc+1}});
        }
        return ans[n-1][n-1];
    }
};
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