<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 bfs(vector<vector<int>>&vis, vector<vector<int>>&grid, int val){
        queue<pair<int,int>> q;
        int n = grid.size();
        int m = grid[0].size();

        if(grid[0][0] < val){
            return;
        }
        q.push({0,0});
             int dx[]= {-1,0,1,0};
        int dy[]= {0,1,0,-1};
        vis[0][0]= 1;
        while(!q.empty()){
         pair<int ,int > tt = q.front();
         q.pop();

         for(int k = 0 ; k<4 ; k++){
            int xx = tt.first + dx[k];
            int yy = tt.second + dy[k];

            if(xx>=0 && yy>=0 && xx<n && yy<m && grid[xx][yy]>=val ){
                if(!vis[xx][yy]){
                    vis[xx][yy] = 1;
                    q.push({xx,yy});
                }
            }
         }

        }
    }

    bool check(int val, vector<vector<int>>&grid){

        vector<vector<int>>vis(grid.size() , vector<int>(grid[0].size() , 0));
        bfs(vis,grid, val);

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


        vector<vector<int>>dis1(grid.size() , vector<int>(grid[0].size() , 1e9));

    queue<pair<int,int>>q;
    int dx[] = {-1, 0, 1 ,0};
    int dy[] = {0,1,0 ,-1};
    int n = grid.size();
    int m = grid[0].size();
    for(int i = 0 ; i<n ; i++){
        for(int j = 0 ; j<m ; j++){
            if(grid[i][j]){            dis1[i][j] = 0;
            q.push({i,j});
            }
        }
    }

    while(!q.empty()){
        pair<int, int> tt = q.front();
        q.pop();

        for(int k = 0 ; k<4 ; k++){
            int xx = tt.first + dx[k];
            int yy = tt.second + dy[k];

                if(xx>=0 && yy>=0 && xx<n && yy<m && dis1[xx][yy] >  1 + dis1[tt.first][tt.second]){
                    q.push({xx, yy});
                    dis1[xx][yy] = 1+ dis1[tt.first][tt.second];
                }
        }
    }

    int lo = 0 ;
    int hi = 1e9;
        int ans ;
    while(lo<=hi){
        int mid = lo + (hi-lo)/2;

        if(check(mid,dis1)){
            ans = mid;
            lo = mid+1;
        }else{
            hi = mid -1;
        }
    }
    return ans;

    }
 
};
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