<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>

struct Point {
    int r, c;
};

struct Node {
    Point point;
    int score;

    bool operator<(const Node& other) const {
        return score < other.score;
    }
    bool operator>(const Node& other) const {
        return score > other.score;
    }
};

class Solution {
public:
    constexpr inline bool in_bounds(int r, int c, int rows, int cols) const noexcept {
        return r >= 0 && r < rows && c >= 0 && c < cols;
    }

    int maximumSafenessFactor(vector<vector<int>>& grid) {
        const int ROWS = grid.size();
        const int COLS = grid.at(0).size();
        
        const int DR[] = {-1, 1, 0, 0};
        const int DC[] = {0, 0, 1, -1};
        

        // {Point, bfs depth}
        queue<Point> q;
        vector<vector<int>> safeness(ROWS, vector<int>(COLS, -1));
        for(int r = 0; r < ROWS; r++) {
            for(int c = 0; c < COLS; c++) {
                if (grid[r][c] == 1) {
                    q.push({r, c});
                    safeness[r][c] = 0;
                }
            }
        }

        // safeness[r][c] is the minimum distaince from (r, c) to any thief 
        // on the grid
        while(!q.empty()) {
            auto [r, c] = q.front(); q.pop();

            for(int i = 0; i < 4; i++) {
                int new_r = r + DR[i];
                int new_c = c + DC[i];

                if (in_bounds(new_r, new_c, ROWS, COLS) && safeness[new_r][new_c] == -1) {
                    safeness[new_r][new_c] = safeness[r][c] + 1;
                    q.push({new_r, new_c});
                }
            }
        }

        for(int r = 0; r < ROWS; r++) {
            for(int c = 0; c < COLS; c++) {
                cout << safeness[r][c];
            }
            cout << endl;
        }
        cout << "----------" << endl;


        // N^2 log N
        // want max heap
        priority_queue<Node, vector<Node>, less<Node>> pq{};
        vector<vector<bool>> visited(ROWS, vector<bool>(COLS, false));

        pq.push(Node{ Point{0, 0}, safeness[0][0]});

        while(!pq.empty()) {
            Node n = pq.top(); pq.pop();
            const int r = n.point.r;
            const int c = n.point.c;
            const int score = n.score;

            if (r == ROWS - 1 && c == COLS - 1) return score;

            for(int i = 0; i < 4; i++) {
                int new_r = r + DR[i];
                int new_c = c + DC[i];

                if (in_bounds(new_r, new_c, ROWS, COLS) && !visited[new_r][new_c]) {
                    visited[new_r][new_c] = true;

                    int new_score = min(score, safeness[new_r][new_c]);
                    pq.push(Node{Point{new_r, new_c}, new_score});
                }
            }
        }

        return -1;
    }
};
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