<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 solve(vector<vector<int>> &grid)
    {
        int n = grid.size();
        int m = grid[0].size();

        vector<vector<int>> safeness(n, vector<int>(m, INT_MIN));
        safeness[0][0] = grid[0][0];

        priority_queue<vector<int>> pq;
        pq.push({grid[0][0], 0, 0});

        while(!pq.empty())
        {
            vector<int> tmp = pq.top();
            pq.pop();

            int currentSafeness = tmp[0];
            int i = tmp[1], j = tmp[2];

            if(i == n-1 && j == m-1)
                return currentSafeness;

            vector<vector<int>> dir = {{0, 1}, {1, 0}, {-1, 0}, {0, -1}};

            for(int d=0; d<4; d++)
            {
                int x = i + dir[d][0];
                int y = j + dir[d][1];

                if(x >= 0 && y >= 0 && x < n && y < m)
                {
                    int newSafeness = min(grid[x][y], currentSafeness);
                    if(newSafeness > safeness[x][y])
                    {
                        safeness[x][y] = newSafeness;
                        pq.push({newSafeness, x, y});
                    }
                }
            }
        }

        return -1;
    }

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

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

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

            int i = tmp.first, j = tmp.second;

            if((i-1) >= 0 && grid[i-1][j] > 1 + grid[i][j])
                grid[i-1][j] = grid[i][j] + 1, q.push({i-1, j});

            if((j-1) >= 0 && grid[i][j-1] > grid[i][j] + 1)
                grid[i][j-1] = grid[i][j] + 1, q.push({i, j-1});

            if((i+1) < n && grid[i+1][j] > grid[i][j] + 1)
                grid[i+1][j] = grid[i][j] + 1, q.push({i+1, j});

            if((j+1) < m && grid[i][j+1] > grid[i][j] + 1)
                grid[i][j+1] = grid[i][j] + 1, q.push({i, j+1});
        }

        vector<vector<bool>> vis(n, vector<bool>(m, false));
        int ans = solve(grid);
        return ans;
    }
};
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