<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 DisjointSet{
private:
       vector<int> par;
       vector<int> size;
public:
       DisjointSet(int n){
          size.resize(n+1,1);
          for(int i=0;i<=n;i++){
            par.push_back(i);
          }
       }

       int findUpar(int a){
        if(par[a]==a) return a;
        return par[a]=findUpar(a);
       }

       void Union(int a, int b){
        int par1=findUpar(a);
        int par2=findUpar(b);
        if(par1==par2) return;
        if(size[par1]>size[par2]){
            par[par2]=par2;
            size[par1]+=size[par2];
        }
        else{
            par[par1]=par2;
            size[par2]+=size[par1];
        }
       }
};

void bfs(int i,int j,vector<vector<int>>&grid,vector<vector<int>>&scores){
    vector<int> drow={1,-1,0,0};
    vector<int> dcol={0,0,1,-1};
    scores[i][j]=0;
    queue<pair<pair<int,int>,int>> q;
    q.push({{i,j},0});
    while(!q.empty()){
        int r=q.front().first.first;
        int c=q.front().first.second;
        int d=q.front().second;
        q.pop();
        for(int k=0;k<4;k++){
            int nr=r+drow[k];
            int nc=c+dcol[k];
            if(nr>=0 && nc>=0 && nr<grid.size() && nc<grid.size()){
                if(d+1<scores[nr][nc]){
                    scores[nr][nc]=d+1;
                    q.push({{nr,nc},d+1});
                }
            }
        }


    }
}
class Solution {
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n=grid.size();
        if(grid[0][0]==1 || grid[n-1][n-1]==1) return 0;
        vector<vector<int>> scores(n,vector<int>(n,1e8));
        DisjointSet ds(n*n-1);
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    bfs(i,j,grid,scores);
                }
            }
        }
        vector<vector<int>>v(n,vector<int>(n,-1));
        v[0][0]=scores[0][0];
        priority_queue<pair<int,pair<int,int>>> pq;
        pq.push({scores[0][0],{0,0}});
        vector<int> drow={1,-1,0,0};
       
    vector<int> dcol={0,0,1,-1};
        while(!pq.empty()){
            int dis=pq.top().first;
            int r=pq.top().second.first;
            int c=pq.top().second.second;
            pq.pop();
            for(int k=0;k<4;k++){
            int nr=r+drow[k];
            int nc=c+dcol[k];
            if(nr>=0 && nc>=0 && nr<grid.size() && nc<grid.size() && grid[nr][nc]==0){
               int m=min(dis,scores[nr][nc]);
               if(m>v[nr][nc]){
               
                v[nr][nc]=m;
                pq.push({m,{nr,nc}});
               }
            }
        }


        }
        if(v[n-1][n-1]==-1) return 0;
        return v[n-1][n-1];

    }
};
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