<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) {
        queue<vector<int>> q;
        int n = grid.size();
        vector<vector<bool>> seen(n, vector<bool>(n));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    q.push({i, j, 0});
                    grid[i][j] = 0;
                    seen[i][j] = true;
                }
            }
        }
        vector<vector<int>> dir = {{0,1}, {1,0}, {0,-1}, {-1,0}};
        while (!q.empty()) {
            int x = q.front()[0];
            int y = q.front()[1];
            int c = q.front()[2];
            q.pop();
            for (const auto& d : dir) {
                if (x + d[0] < 0 || x + d[0] >= n) continue;
                if (y + d[1] < 0 || y + d[1] >= n) continue;
                if (seen[x + d[0]][y + d[1]]) continue;
                seen[x + d[0]][y + d[1]] = true;
                grid[x + d[0]][y + d[1]] = c+1;
                q.push({x + d[0], y + d[1], c+1});
            }
        }
        auto comp = [] (vector<int>& a, vector<int>& b) { return a[0] < b[0]; };
    priority_queue<vector<int>, vector<vector<int>>, decltype(comp)> pq(comp);

        //insert frist
        //keep on getting the highest one available, once end is reached you have the score
        pq.push({grid[0][0], 0, 0});
        seen[0][0] = false;
        int current_score = grid[0][0];
        while (!pq.empty()) {
            int x = pq.top()[1];
            int y = pq.top()[2];
            current_score = min(current_score, grid[x][y]);
            pq.pop();
            for (const auto& d : dir) {
                if (x + d[0] < 0 || x + d[0] >= n) continue;
                if (y + d[1] < 0 || y + d[1] >= n) continue;
                if (!seen[x + d[0]][y + d[1]]) continue;
                seen[x + d[0]][y + d[1]] = false;
                pq.push({grid[x + d[0]][y + d[1]], x + d[0], y + d[1]});
                if ((x + d[0] == n-1) && (y + d[1] == n-1)) {
                    return min(current_score, grid[n-1][n-1]);
                }
            }
        }

        return current_score;
    }
};
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