<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;
    vector<pair<int, int>> index = {{1,0}, {-1,0}, {0,1}, {0,-1}};
    vector<vector<int>> vis;
    void fillDist(vector<vector<int>>& grid) {
        int k=0; n=grid.size();
        vis.resize(n,vector<int>(n,0));
        queue<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({i,j}); vis[i][j]=1;}
            }
        }

        while(!q.empty()) {
            int sz = q.size();
            while(sz--) {
                auto [x,y] = q.front(); 
                q.pop();
                grid[x][y]=k;
                for(auto [i,j]: index){
                    if(x+i<n && y+j<n && x+i>=0 && y+j>=0 && vis[x+i][y+j]==0){q.push({x+i, y+j}); vis[x+i][y+j]=1;}
                }
            }
            k++;
        }
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        fillDist(grid);
        priority_queue<vector<int>> pq;
        int ans = 0; n=grid.size();
        vis.clear();
        vis.resize(n,vector<int>(n,0));
        // for(int i=0;i<n;i++){
        //     for (int j=0;j<n;j++){
        //         cout<<grid[i][j]<<" ";
        //     }
        //     cout<<endl;
        // }
        // cout<<endl;

        pq.push({grid[0][0], 0, 0});
        vis[0][0]=1;
        while(!pq.empty()){
            vector<int> vec = pq.top();
            pq.pop();
            if(vec[1]==n-1 && vec[2]==n-1)ans = max(ans, vec[0]);
            for(auto [i,j]: index){
                if(vec[1]+i<n && vec[2]+j<n && vec[1]+i>=0 && vec[2]+j>=0 && vis[vec[1]+i][vec[2]+j]==0) {
                    pq.push({min(vec[0], grid[vec[1]+i][vec[2]+j]), vec[1]+i, vec[2]+j});
                    vis[vec[1]+i][vec[2]+j]=1;
                }
            }
        }
        return ans;
    }
};
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