<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 dx[4] = {0, 1, -1, 0};
    int dy[4] = {1, 0, 0, -1};
    bool valid(int x, int y, vector<vector<int>>& grid){
        return x > -1 && y > -1 && x < grid.size() && y < grid[0].size();
    }
    void bfs(int x, vector<vector<int>>& grid){
        int n = grid.size(), m = grid[0].size();
        queue<pair<int, int>>qu;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < m; j++){
                if(grid[i][j]){
                    qu.push({i, j});
                    grid[i][j] = 0;
                }
                else{
                    grid[i][j] = -1;
                }
            }
        }
        while(qu.size()){
            auto [i, j] = qu.front(); qu.pop();
            for(int k = 0; k < 4; k++){
                int x = i + dx[k]; 
                int y = j + dy[k];
                if(valid(x, y, grid) && grid[x][y] == -1){
                    grid[x][y] = grid[i][j] + 1;
                    qu.push({x, y});
                }
            }
        }
    }
    int id;
    vector<vector<int>>vis;
    bool dfs(int i, int j, int &md, vector<vector<int>>& grid){
        int n = grid.size(), m = grid[0].size();
        if(!valid(i, j, grid) || vis[i][j] == id || grid[i][j] < md)
            return 0;
        if(i == n - 1 && j == m - 1)
            return 1;
        vis[i][j] = id;
        bool ret = 0;
        for(int k = 0; k < 4; k++){
            int x = i + dx[k];
            int y = j + dy[k];
            ret = (ret || dfs(x, y, md, grid));
        }
        return ret;
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        bfs(m, grid);
        int l = 0, r = 400;
        vis = vector<vector<int>>(n, vector<int>(m, 0));
        while(l <= r){
            id++;
            int md = (l + r) >> 1;
            if(dfs(0, 0, md, grid))
                l = md + 1;
            else
                r = md - 1; 
        }
        return l - 1;
    }
};
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