<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:
bool possible(int mid,vector<vector<int>>&dis){
    queue<pair<int,pair<int,int>>>q;
    if(dis[0][0]<mid)
    return false;
    q.push({dis[0][0],{0,0}});
    int n=dis.size();
    vector<vector<int>>vis(n,vector<int>(n));
    int dr[]={-1,1,0,0};
    int dc[]={0,0,-1,1};
    vis[0][0]=1;
    while(!q.empty()){
        auto d=q.front().first;
        int row=q.front().second.first;
        int col=q.front().second.second;
        q.pop();
        if(row==n-1 && col==n-1)
        return true;
        for(int i=0;i<4;i++){
            int nrow=row+dr[i];
            int ncol=col+dc[i];
            if(nrow>=0 && ncol>=0 && nrow<n && ncol<n && dis[nrow][ncol]>=mid && !vis[nrow][ncol]){
                q.push({dis[nrow][ncol],{nrow,ncol}});
                vis[nrow][ncol]=1;
            }
        }
    }
    return false;
}
    int maximumSafenessFactor(vector<vector<int>>& grid) {
       int m=grid.size();
       int n=grid[0].size();
       vector<vector<int>>dis(m,vector<int>(n,1e9));
       queue<pair<int,pair<int,int>>>q;
       for(int i=0;i<n;i++){
        for(int j=0;j<n;j++){
            if(grid[i][j]==1){
                q.push({0,{i,j}});
                dis[i][j]=0;
            }
        }
       }
       int dr[]={-1,1,0,0};
       int dc[]={0,0,-1,1};
       while(!q.empty()){
        auto d=q.front().first;
        int row=q.front().second.first;
        int col=q.front().second.second;
        q.pop();
        for(int i=0;i<4;i++){
            int nrow=row+dr[i];
            int ncol=col+dc[i];
            if(nrow>=0 && ncol>=0 && nrow<n && ncol<n && dis[nrow][ncol]>dis[row][col]+1){
                dis[nrow][ncol]=dis[row][col]+1;
                q.push({d+1,{nrow,ncol}});
            }
        }
       }
    int low=0;
    int high=1e9;
    int ans=-1;
    while(low<=high){
        int mid=low+(high-low)/2;
        if(possible(mid,dis)){
            ans=mid;
            low=mid+1;
        }
        else
        high=mid-1;
    }
       return ans;
    }
};
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