<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 maximumSafenessFactor(vector<vector<int>>& grid) {
        const int dirs[][2] = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
        const int M = grid.size();
        const int N = M == 0 ? 0 : grid[0].size();
        vector< vector<int> > safeness(M, vector<int>(N, -1));
        queue< pair<int, int> > que;
        int safe = 1;
        vector< vector<int> > visited(M, vector<int>(N));
        priority_queue< vector<int> > maxHeap;

        for (int i = 0; i < M; ++i) {
            for (int j = 0; j < N; ++j) {
                if (grid[i][j] == 1) {
                    que.emplace(i, j);
                    safeness[i][j] = 0;
                }
            }
        }

        while (!que.empty()) {
            for (int i = que.size(); i > 0; --i) {
                auto p = que.front();
                auto x = p.first;
                auto y = p.second;

                que.pop();

                for (const auto& dir : dirs) {
                    auto dx = x + dir[0];
                    auto dy = y + dir[1];

                    if (dx >= 0 && dx < M && dy >= 0 && dy < N && safeness[dx][dy] == -1) {
                        safeness[dx][dy] = safe;
                        que.emplace(dx, dy);
                    }
                }
            }

            ++safe;
        }

        maxHeap.push({safeness[0][0], 0, 0});

        while (!maxHeap.empty()) {
            auto v = maxHeap.top();
            auto d = v[0];
            auto x = v[1];
            auto y = v[2];

            maxHeap.pop();

            if (x + 1 == M && y + 1 == N) {
                return d;
            }

            if (visited[x][y] != 0) {
                continue;
            }
            visited[x][y] = 1;

            for (const auto& dir : dirs) {
                auto dx = x + dir[0];
                auto dy = y + dir[1];

                if (dx >= 0 && dx < M && dy >= 0 && dy < N && visited[dx][dy] == 0) {
                    maxHeap.push({min(d, safeness[dx][dy]), dx, dy});
                }
            }
        }

        return -1;
    }
};
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