<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:
    vector<int> dx {0,1,0,-1};
    vector<int> dy {1,0,-1,0};
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size() , m = grid[0].size() ;
        if (grid[0][0] || grid[n-1][m-1]) return 0 ;
        queue<pair<int,int>> q;
        for(int i = 0 ; i < n ; i++) 
            for(int j = 0 ; j < m ; j++)    
                if (grid[i][j])
                    q.push({i,j}) ;
        while(q.size()){
            int sz = q.size() ;
            while (sz--){
                auto [x,y] = q.front() ; q.pop() ;
                for(int i = 0 ; i < 4; i++){
                    int nx = x + dx[i] , ny = y + dy[i] ;
                    if (nx >= 0 && ny >= 0 && nx < n && ny < m && !grid[nx][ny])
                    {
                        grid[nx][ny] = grid[x][y] + 1 ; 
                        q.push({nx,ny}) ;
                    }
                }        
            }
        }
        for(auto  I: grid) {
            for(auto j : I) cout << j <<" " ;
            cout << endl;
        }
        auto fx = [&] (int k){
            q = queue<pair<int,int>> () ;
            vector<vector<int>> vis (n , vector<int> (m)) ;
            if (grid[0][0]-1 < k) return 0 ;
            q.push({0,0});
            vis[0][0] = 1 ;
            while(q.size()){
                auto [x,y] = q.front() ; q.pop() ;
                if (x == n-1 && y == m-1) return 1;
                
                for(int i = 0 ; i < 4; i++){
                    int nx = x + dx[i] , ny = y + dy[i] ;
                    if (nx >= 0 && ny >= 0 && nx < n && ny < m && !vis[nx][ny] && grid[nx][ny]-1 >= k)
                    {
                        q.push({nx,ny}) ;
                        vis[nx][ny] = 1 ; 
                    }
                }        
            }
            return 0 ;
        };

        int lo = 0 , hi = 400 ;
        while(lo <= hi){
            int mid = (lo + hi)>>1;
            if (fx(mid)) lo = mid+1 ;
            else hi = mid-1 ;
        }
        return hi ;
    }
};
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