<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 dx[4] = {1,-1,0,0};
    int dy[4] = {0,0,-1,1};
    bool bfs(vector<vector<int>>&grid,vector<vector<int>>&dist,int x){
        int n = grid.size();
        vector<vector<int>>vis(n,vector<int>(n));
        vis[0][0] = 1;
        queue<pair<int,int>>q;
        q.push({0,0});
        if(dist[0][0]<x) return false;
        while(!q.empty()){
            auto cur = q.front();
            q.pop();
            int r = cur.first;
            int c = cur.second;
            if(r==n-1 && c==n-1){
                return true;
            }
            for(int k=0;k<4;k++){
                    int nr = dx[k]+r;
                    int nc = dy[k]+c;
                    if(nr>=0 && nr<n && nc>=0 && nc<n && !vis[nr][nc] && dist[nr][nc]>=x){
                        vis[nr][nc] = 1;
                        q.push({nr,nc});
                    }
            }
        }
        return false;

    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<pair<int,pair<int,int>>>q;
        vector<vector<int>>vis(n,vector<int>(n));
        vector<vector<int>>dist(n,vector<int>(n,1e9));
        for(int i = 0;i<n;i++){
            for(int j = 0;j<n;j++){
                if(grid[i][j]==0) continue;
                dist[i][j] = 0;
                vis[i][j]=1;
                q.push({dist[i][j],{i,j}});
            }
        }
        while(!q.empty()){
            auto cur = q.front();
            q.pop();
            int dis = cur.first;
            int r = cur.second.first;
            int c = cur.second.second;
            for(int k=0;k<4;k++){
                    int nr = dx[k]+r;
                    int nc = dy[k]+c;
                    if(nr>=0 && nr<n && nc>=0 && nc<n && !vis[nr][nc]){
                        vis[nr][nc] = 1;
                        dist[nr][nc] = 1+dist[r][c];
                        q.push({dist[nr][nc],{nr,nc}});
                    }
            }

        }
        int lo = 0 , hi = n*n;
        int ans = 0;
        while(hi-lo>1){
            int mid = (lo+hi)/2;
            if(bfs(grid,dist,mid)){
                ans = max(ans,mid);
                lo = mid;
            }
            else{
                hi = mid-1;
            }
        }
        if(bfs(grid,dist,hi)) ans = max(ans,hi);
        return ans;


    }
};
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