<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 INF = 1e5;
    vector<int> dr = {-1, 1, 0, 0};
    vector<int> dc = {0, 0, 1, -1};

    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> costs(n, vector<int>(n, INF));
        vector<vector<int>> dists(n, vector<int>(n));
        vector<pair<int,int>> thieves;
        queue<vector<int>> Q;
        priority_queue<vector<int>> PQ;

        for (int r = 0; r < n; r++)
            for (int c = 0; c < n; c++)
                if (grid[r][c])
                    thieves.push_back({r,c});

        for (auto p : thieves) {
            Q.push({p.first, p.second, 0});
            costs[p.first][p.second] = 0;
        }
        
        while (!Q.empty()) {
            auto p = Q.front(); Q.pop();
            int r = p[0], c = p[1], d = p[2];
            for (int i = 0; i < 4; i++) {
                int nr = r + dr[i];
                int nc = c + dc[i];
                if (in_bounds(nr, nc, n) and costs[nr][nc] == INF) {
                    costs[nr][nc] = d + 1;
                    Q.push({nr, nc, d + 1});
                }
            }
        }

        dists[0][0] = costs[0][0];
        PQ.push({dists[0][0], 0, 0});
        while (!PQ.empty()) {
            auto p = PQ.top(); PQ.pop();
            int r = p[1], c = p[2];
            for (int i = 0; i < 4; i++) {
                int nr = r + dr[i];
                int nc = c + dc[i];

                if (!in_bounds(nr, nc, n))
                    continue;

                int move_cost = min(dists[r][c], costs[nr][nc]);
                if (move_cost > dists[nr][nc]) {
                    dists[nr][nc] = move_cost;
                    PQ.push({move_cost, nr, nc});
                }
            }
        }

        return dists[n-1][n-1];
    }

    bool in_bounds(int r, int c, int n) {
        return r >= 0 and r < n and c >= 0 and c < n; 
    }
};
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