<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:
    int n,m;
    vector<vector<int>>dist;
    vector<vector<int>>dir{ {-1,0} , {0,1} , {1,0} , {0,-1}};
    
    void preComputeMinDist_bfs(vector<vector<int>>&grid){
        
         queue<vector<int>>q;
         vector<vector<bool>>vis(n,vector<bool>(m,false));
    
         for(int i = 0 ; i < grid.size() ; i++){
             for(int j = 0 ; j < grid[i].size() ; j++){
                 if(grid[i][j] == 1){
                      vis[i][j] = true;
                      q.push({i,j,0});
                      dist[i][j] = 0;
                 }
             }
         }

         while(!q.empty()){
              auto v = q.front();
              q.pop();

              int row = v[0];
              int col = v[1];
              int d   = v[2];

              for(int i = 0 ; i < dir.size() ; i++){
                  int r = dir[i][0];
                  int c = dir[i][1];
                  if(isValid(row+r,col+c) == true and vis[row+r][col+c] == false){
                       vis[row+r][col+c] = true;
                       dist[row+r][col+c] = d+1;
                       q.push({row+r,col+c,d+1});
                  }
              }
         }
         //print(dist);
         return;

    }
    void print(vector<vector<int>>&arr){
        for(int i = 0 ; i < arr.size() ; i++){
            for(int j = 0 ; j < arr[i].size() ; j++){
                cout<<arr[i][j]<<" ";
            }
            cout<<endl;
        }
        cout<<endl<<endl;

        return;
    }
    
    int maximumSafenessFactor(vector<vector<int>>& grid) {
       if(grid[0][0] == 1 or grid[grid.size()-1][grid[0].size()-1] == 1) 
        return 0;

        n = grid.size();
        m = grid[0].size();

        dist.resize(n,vector<int>(m,INT_MAX));  //to keep the min dis of cell from theif
        preComputeMinDist_bfs(grid);

        int start = 0;
        int end   = INT_MAX;
        int ans = end;
        
        while(start <= end){
            int mid = (0LL + start + end)/(1LL * 2);
            if(bfs(mid) == true){
                 ans = mid;
                 start = mid+1;   //we will try to maximize distance 
            }
            else{
                 end = mid -1;  //we will try to minimize distance
            }
            //cout<<"ans = "<<ans<<endl<<endl<<endl;
        }
        return ans;
    }
    //dist[row][col] = x
    //x is the minimum distance from any theive in the (grid) to the cell row col in the (path)
    bool bfs(int &mid){
         //cout<<"mid = "<<mid<<endl<<endl<<endl;

         queue<pair<int,int>>q;
         vector<vector<bool>>vis(n,vector<bool>(m,false));
         
         if(dist[0][0] >= mid){
             q.push({0,0});
             vis[0][0] = true;
         }
         else return false;

         while(!q.empty()){

              auto p = q.front(); q.pop();
              int row = p.first;
              int col = p.second;

              if(row == n-1 and col == m-1)
              return true;
                
              for(int i = 0 ; i < dir.size() ; i++){
                  int r = dir[i][0];
                  int c = dir[i][1];
                  //cout<<"row = "<<row<<" "<<"col = "<<col<<endl;

                  if(isValid(row+r,col+c) == true and vis[row+r][col+c] == false and dist[row+r][col+c] >= mid){
                      //cout<<"row+r = "<<row+r<<" "<<"col+c = "<<col+c<<endl;

                      q.push({row+r,col+c});
                      vis[row+r][col+c] = true;
                  }
              }   
              
         }
         return false;
    }
    bool isValid(int row , int col){
        if(row >= 0 and row < n and col >=0 and col < m){
            return true;
        }
        return false;
    }
};
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