<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:
    
    bool safe(int r, int c, int n) {
        return(r>=0 && c>=0 && r<n && c<n);
    }
    
    vector<int> x = {-1, 1, 0, 0};
    vector<int> y = {0, 0, -1, 1};
    
    class Comp {
        public:
            bool operator() (vector<int> &a, vector<int> &b) {
                return a[0]<b[0];
            }
    };
    
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        
        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(make_pair(i, j));
                    grid[i][j] = 0;
                }
                else {
                    grid[i][j] = 1e4;
                }
            }
        }
        int step = 1;
        while(!q.empty()) {
            int l = q.size();
            while(l--) {
                int row = q.front().first;
                int col = q.front().second; 
                q.pop();
                for(int i=0;i<4;i++) {
                    int xp = row+x[i];
                    int yp = col+y[i];
                    if(safe(xp, yp, n) && grid[xp][yp]==1e4) {
                        grid[xp][yp] = step;
                        q.push(make_pair(xp, yp));
                    }
                }
            }
            step++;
        }
        
        // for(int i=0;i<n;i++) {
        //     for(int j=0;j<n;j++) cout << grid[i][j] << " ";
        //     cout << endl;
        // }
        // cout << "--" << endl;
        
        // done with filling
        vector<vector<int>> dp(n, vector<int> (n, -1));
        dp[0][0] = grid[0][0];
        priority_queue<vector<int>, vector<vector<int>>, Comp> pq;
        pq.push({dp[0][0], 0, 0});
        
        while(!pq.empty()) {
            int row = pq.top()[1];
            int col = pq.top()[2];
            int dist = pq.top()[0];
            // cout << row << " " << col << " " << dist << endl;
            pq.pop();
            for(int i=0;i<4;i++) {
                int xp = row+x[i];
                int yp = col+y[i];
                if(safe(xp, yp, n)) {
                    if(dp[xp][yp]==-1) {
                        dp[xp][yp] = min(dist, grid[xp][yp]);
                        pq.push({dp[xp][yp], xp, yp});
                    }
                    // else {
                    //     if(dp[xp][yp]<min(dist, grid[xp][yp])) {
                    //         dp[xp][yp] = min(dist, grid[xp][yp]);
                    //         pq.push({dp[xp][yp], xp, yp});
                    //     }
                    // }
                }
            }
        }
        
        // cout << "--" << endl;
        // for(int i=0;i<n;i++) {
        //     for(int j=0;j<n;j++) cout << dp[i][j] << " ";
        //     cout << endl;
        // }
        return dp[n-1][n-1];
        
    }
};
217821