<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 {

    int dx[4]={1, -1, 0, 0};
    int dy[4]={0, 0, 1, -1};
    
    bool bfs(vector<vector<int>> &dist, int sf)
    {
        if(dist[0][0]<sf) return false;
        int n=dist.size();
        queue<pair<int, int>> q;
        vector<vector<bool>> vis(n, vector<bool> (n, false));
        q.push({0, 0});
        vis[0][0]=true;
        while(!q.empty())
        {
            int x=q.front().first;
            int y=q.front().second;
            q.pop();
            if(x==n-1 && y==n-1) return true;
            for(int k=0; k<4; k++)
            {
                int nx=x+dx[k];
                int ny=y+dy[k];
                if(nx>=0 && nx<n && ny>=0 && ny<n && dist[nx][ny]>=sf && !vis[nx][ny])
                {
                    q.push({nx, ny});
                    vis[nx][ny]=true;
                }
            }
        }
        return false;
    }
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) 
    {
        int n=grid.size();
        queue<pair<int, int>> q;
        vector<vector<bool>> vis(n, vector<bool> (n, false));
        vector<vector<int>> dist(n, vector<int> (n, INT_MAX));
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++) 
            {
                if(grid[i][j]==1) 
                {
                    dist[i][j]=0;
                    q.push({i, j});
                    vis[i][j]=true;
                }
            }
        }
        while(!q.empty())
        {
            int sz=q.size();
            while(sz--)
            {
                int x=q.front().first, y=q.front().second;
                q.pop();
                for(int k=0; k<4; k++)
                {
                    int nx=x+dx[k];
                    int ny=y+dy[k];
                    if(nx>=0 && nx<n && ny>=0 && ny<n && !vis[nx][ny])
                    {
                        q.push({nx, ny});
                        vis[nx][ny]=true;
                        dist[nx][ny]=1+dist[x][y];
                    }
                }
            }
        }
        int low=0, high=1e9, ans=0;
        while(low<=high)
        {
            int mid=(low+high)/2;
            if(bfs(dist, mid))
            {
                ans=mid;
                low=mid+1;
            }
            else high=mid-1;
        }
        return ans;
    }
};
199091