<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 {
private:
    vector<int> DIR = {0, 1, 0, -1, 0};
    int m = 0;
    int n = 0;
    // is there a valid path of safeness factor d from grid[r][c] to grid[m - 1][n - 1]
    // safeness factor d: every cell on the path >= d
    bool validPath(int d, int r, int c, vector<vector<int>>& distance, vector<vector<bool>>& visited) {
        if (distance[r][c] < d) {
            return false;
        }
        if (r == m - 1 && c == n - 1) {
            return true;
        }
        visited[r][c] = true;
        for (int i = 0; i < 4; ++i) {
            int nr = r + DIR[i], nc = c + DIR[i+1];
            if (nr < 0 || nr == m || nc < 0 || nc == n || visited[nr][nc]) continue;
            if (validPath(d, nr, nc, distance, visited)) {
                return true;
            }
        }
        return false;
    }
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        m = grid.size();
        n = grid[0].size();
        if (grid[0][0] == 1 || grid[m - 1][n - 1] == 1) {
            return 0;
        }
        // distance[i][j] is the smallest distance from grid[i][j] to thief
        vector<vector<int>> distance(m, vector<int>(n, -1));
        queue<pair<int, int>> q;
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (grid[i][j] == 1) {
                    q.push({i, j});
                    distance[i][j] = 0;
                }
            }
        }
        int d = 1;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; ++i) {
                pair<int, int> p = q.front();
                q.pop();
                int r = p.first, c = p.second;
                for (int i = 0; i < 4; ++i) {
                    int nr = r + DIR[i], nc = c + DIR[i+1];
                    if (nr < 0 || nr == m || nc < 0 || nc == n || distance[nr][nc] != -1) continue;
                    distance[nr][nc] = d;
                    q.push({nr, nc});
                }
            }
            d++;
        }
        int left = 0, right = d - 1;
        int ans = 0;
        while (left <= right) {
            int mid = left + (right - left) / 2;
            cout << mid;
            vector<vector<bool>> visited(m, vector<bool>(n, false));
            if (validPath(mid, 0, 0, distance, visited)) {
                left = mid + 1;
                ans = mid;
                cout << " is valid" << std::endl;
            } else {
                cout << endl;
                right = mid - 1;
            }
        }
        return ans;
    }
};
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