<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 maximumSafenessFactor(vector<vector<int>>& grid) {
        
        
        queue<vector<int>> q;
        int n = grid.size();

        if(grid[n-1][n-1] == 1)
        return 0;
        vector<vector<int>> dist(n, vector<int>(n,-1));

        for(int i=0; i< n; i++) {
            for(int j=0; j< n; j++) {
                if(grid[i][j])
                q.push({i,j, 0}), dist[i][j] = 0;
            }
        }

        int dx[4] = {-1,1,0,0};
        int dy[4] = {0,0,1,-1};
        int newx, newy;
        int ans = INT_MAX;

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

            for(int k=0; k<4; k++) {
                newx = t[0] + dx[k];
                newy = t[1] + dy[k];

                if(newx >=0 && newx < n && newy >= 0 && newy < n) {
                    if(dist[newx][newy] == -1 )
                    {
                        dist[newx][newy] = 1 + t[2];
                        q.push({newx,newy, dist[newx][newy]});
                    }
                    else {
                        dist[newx][newy] = min(dist[newx][newy], 
                        1 + t[2]);
                    }
                   // cout<<newx<<" "<<newy<<" "<<dist[newx][newy]<<endl;
                  //  ans = min(ans, dist[newx][newy]);
                }
            }
        }

        q.push({0,0});
        grid[0][0] = -1;
        ans = dist[0][0];
        priority_queue<vector<int>> pq;
        pq.push({dist[0][0], 0,0});

        while(!pq.empty()) {
            auto t = pq.top();
            pq.pop();
            ans = min(ans, t[0]);
          //  cout<<t[0] <<" "<<t[1]<<" "<<t[2]<<endl;

            if(t[1] == n-1 && t[2] == n-1)
            return ans;

            for(int k=0; k<4; k++) {
                newx = t[1] + dx[k];
                newy = t[2] + dy[k];
                if(newx >=0 && newx < n && newy >= 0 && newy < n && grid[newx][newy] != -1) {
                    pq.push({dist[newx][newy],newx, newy});
                    grid[newx][newy] = -1;
                   // ans = min(dist[newx][newy], ans);
                }

            }


        }

        return ans;
        
    }
};
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