<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:
    inline bool isValid(int i, int j, int m, int n) {
        return i >= 0 && i < m && j >= 0 && j < n;
    }
    void updateSafeFactor(int i, int j, vector<vector<int>>& grid, vector<vector<int>> &safeFactor) {
        int m = grid.size();
        int n = grid[0].size();
        
        safeFactor[i][j] = 0;
        queue<pair<int, pair<int, int>>> pointQueue;
        pointQueue.push({0, {i, j}});
        const vector<pair<int, int>> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
        while (!pointQueue.empty()) {
            auto [curSafeFactor, point] = pointQueue.front();
            auto [x, y] = point;
            // cout << x << " " << y << " " << curSafeFactor << "\n";
            pointQueue.pop();
            for (auto &[dx, dy] : dirs) {
                if (!isValid(x + dx, y + dy, m, n)) {
                    continue;
                }
                if (grid[x+dx][y+dy] == 1) {
                    continue;
                }
                if (curSafeFactor + 1 < safeFactor[x + dx][y + dy]) {
                    safeFactor[x + dx][y + dy] = curSafeFactor + 1;
                    pointQueue.push({curSafeFactor + 1, {x + dx, y + dy}});
                }

            }
        }

    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int m = grid.size();
        int n = grid[0].size();
        int lo = 0;
        int hi = m + n;
        vector<vector<int>> safeFactor(m, vector<int>(n, m+n));

        for (int i = 0; i < m; i ++) {
            for (int j = 0; j < n; j ++) {
                if (grid[i][j] == 1) {
                    updateSafeFactor(i, j, grid, safeFactor);
                }

            }
        }

        priority_queue<pair<int, pair<int, int>>> pointQueue;
        pointQueue.push({safeFactor[0][0], {0, 0}});
        vector<vector<bool>> visited(m, vector<bool>(n, false));
        visited[0][0] = true;
        while (!pointQueue.empty()) {
            auto [curSafeFactor, point] = pointQueue.top();
            auto &[x, y] = point;
            pointQueue.pop();
            if (x == m - 1 && y == n - 1) {
                return curSafeFactor;
            }
            const vector<pair<int, int>> dirs = {{0, 1}, {1, 0}, {0, -1}, {-1, 0}};
            for (auto &[dx, dy] : dirs) {
                if (!isValid(x + dx, y + dy, m, n) || grid[x+dx][y+dy] == 1) {
                    continue;
                }
                if (!visited[x+dx][y+dy]) {
                    pointQueue.push({min(safeFactor[x + dx][y + dy], curSafeFactor), {x + dx, y + dy}});
                    visited[x + dx][y + dy] = true;
                }
            }
        }

        return 0;
    }
};
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