<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]={0,1,0,-1};
    int dy[4]={1,0,-1,0};

    bool possible(int val ,vector<vector<int>>& grid )
    {
        queue<pair<int,int>> q;
        int n=grid.size();
        if(grid[0][0]>=val)
        q.push({0,0});
        vector<vector<int>> vis(n, vector<int>(n,0));
        int x, y;

        while(!q.empty())
        {
            x=q.front().first;
            y=q.front().second;
            q.pop();
            if(vis[x][y]==1)continue;
            vis[x][y]=1;
            if(x==n-1 &&y==n-1)
                return true;

            for(int i=0;i<4;i++)
            {
                int newx=x+dx[i];
                int newy=y+dy[i];
                if(newx<0 || newx>=n || newy<0 || newy>=n)
                    continue;
                if(vis[newx][newy]==0 && grid[newx][newy]>=val)
                { 
                    q.push({newx,newy});
                }
            }
        }
        
        return false;
    }
    int func(vector<vector<int>>& grid)
    {

        int n=grid.size();
        vector<vector<int>> visited(n,vector<int>(n,INT_MAX));
        queue<pair<pair<int,int>,int>> pq;
        for(int i=0;i<n;i++)
        for(int j=0;j<n;j++)
        {
            if(grid[i][j]==1)
            {
                pq.push({{i,j},0});
                visited[i][j]=0;
            }
        }
        while(!pq.empty())
        {
            pair<pair<int,int>,int> temp = pq.front();
            pq.pop();
            int x = temp.first.first;
            int y = temp.first.second;
            int dis = temp.second;
            
            for(int i=0;i<4;i++)
            {
                int newx = dx[i]+x;
                int newy = dy[i]+y;
                if(newx<0 || newx>=n || newy<0 || newy>=n)
                    continue;
                if(visited[newx][newy]>dis+1)
                {
                    visited[newx][newy]=dis+1;
                    pq.push({{newx,newy},dis+1});
                }
            }
        }

        int low = 0;
        int high = 2*(n-1) ;
        int mid;
        while(low<=high)
        {
            
            mid=(low+high)>>1;
            if(possible(mid, visited))
            {
                low=mid+1;
            }
            else 
            {

                high=mid-1;
            }
        }
        return high;
        
    }
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        if(grid[0][0]==1||grid[grid.size()-1][grid.size()-1]==1)return 0;
        return func(grid);
        return 0;
    }
};
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