<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:

    void dfs(int i,int j,vector<vector<int>> &dis, vector<vector<int>> &vis,int mid)
    {
        if(dis[i][j]<mid) return;
        int n=dis.size();
        vis[i][j]=1;
        int dx[4]={0,0,1,-1};
        int dy[4]={1,-1,0,0};
        for(int k=0;k<4;k++)
        {
            int nx=i+dx[k];
            int ny=j+dy[k];
            if(nx>=0&&nx<n&&ny>=0&&ny<n&&!vis[nx][ny]&&dis[nx][ny]>=mid)
                dfs(nx,ny,dis,vis,mid);
        }
    }

    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n=grid.size();
        vector<vector<int>> dis(n,vector<int>(n));
        queue<pair<int,int>> Q;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                dis[i][j]=INT_MAX;
                if(grid[i][j]==1){
                    Q.push({i,j});
                    dis[i][j]=0;
                }
            }
        }
       
        while(!Q.empty())
        {
            auto cur=Q.front();
            int x=cur.first;
            int y=cur.second;
            Q.pop();
            int dx[4]={0,0,1,-1};
            int dy[4]={1,-1,0,0};
            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&&dis[nx][ny]>dis[x][y]+1)
                {
                    Q.push({nx,ny});
                    dis[nx][ny]=dis[x][y]+1;
                }
            }
        }
        // for(int i=0;i<n;i++){
        //     for(int j=0;j<n;j++)
        //         cout<<dis[i][j]<<" ";
        //     cout<<"\n";
        // }
        //BS on answer
        int l=0,r=INT_MAX,ans=r;
        while(l<=r)
        {
            int mid=(l+r)/2;
            vector<vector<int>> vis(n,vector<int>(n,0));
            dfs(0,0,dis,vis,mid);
            if(vis[n-1][n-1]) // all distances should be <= mid
            {
                ans=mid;
                l=mid+1;
            }
            else
                r=mid-1;
        }

        return ans;
    }
};
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