<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:
    bool Check(int i, int j, int n){
        return i < n && j < n && i >= 0 && j >= 0;
    }

    int maximumSafenessFactor(vector<vector<int>>& grid) {
        std::queue<std::vector<int>> q;
        int n = grid.size();
        vector<vector<int>> safety(n, vector<int>(n, INT_MAX));
        for (int i = 0; i < n; i++){
            for (int j = 0; j < n; j++){
                if (grid[i][j] == 1){
                    safety[i][j] = 0;
                    q.push({i, j});
                }
            }
        }
        int x[] = {1, -1, 0, 0};
        int y[] = {0, 0, 1, -1};
        while (!q.empty()){
            int size = q.size();
            for (int k = 0; k < size; k++){
                auto pos = q.front();
                q.pop();
                for (int i = 0; i < 4; i++){
                    int ni = pos[0] + x[i];
                    int nj = pos[1] + y[i];
                    if (Check(ni, nj, n) && safety[ni][nj] > safety[pos[0]][pos[1]] + 1){
                        safety[ni][nj] =  safety[pos[0]][pos[1]] + 1;
                        q.push({ni, nj});
                    }
                }
            }
        }
        vector<vector<int>> visit(n, vector<int>(n, 0));
        priority_queue<pair<int, vector<int>>> pq;
        pq.push({safety[0][0], {0, 0}});
        visit[0][0] = 1;
        while (!pq.empty()){
            auto pos = pq.top();
            pq.pop();
            if (pos.second[0] == n - 1 && pos.second[1] == n - 1){
                return pos.first;
            }
            for (int i = 0; i < 4; i++){
                int ni = pos.second[0] + x[i];
                int nj = pos.second[1] + y[i];
                if (Check(ni, nj, n) && !visit[ni][nj]){
                    int s = min(safety[ni][nj], pos.first);
                    visit[ni][nj] = 1;
                    pq.push({s, {ni, nj}});
                }
            }
        }
        return 0;
    }
};
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