<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:

    const vector<pair<int,int>> directions = {{0,1},{0,-1},{1,0},{-1,0}};

    bool isValid(int x, int y, int n) {
        return x >= 0 && x < n && y >= 0 && y < n;
    }
    
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();

        // Initialize BFS queue with thieves and set their safeness to zero
        queue<pair<int,int>> q;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 0) {
                    grid[i][j] = -1;
                } else {
                    q.emplace(i, j);
                    grid[i][j] = 0;
                }
            }
        }

        // Perform BFS inplace and update the safeness factor of each grid
        while(!q.empty()) {
            auto [i, j] = q.front();
            q.pop();
            int currDist = grid[i][j] + 1;
            for (auto [dx,dy] : directions) {
                if (isValid(i + dx, j + dy, n) && grid[i+dx][j+dy] == -1) {
                    grid[i+dx][j+dy] = currDist;
                    q.emplace(i+dx,j+dy);
                }
            }
        }
        
        // To find max safe path, greedily choose max safeness factor based on neighbors
        typedef pair<int,pair<int,int>> entry;
        priority_queue<entry> pq;
        pq.push({grid[0][0], {0,0}});
        
        while (!pq.empty()) {
            int maxSafeness = pq.top().first;
            auto [i,j] = pq.top().second;
            pq.pop();

            if (grid[i][j] == -1) continue;

            grid[i][j] = -1;
            if (i == n-1 && j == n-1) {
                return maxSafeness;
            }
            
            for (auto [dx,dy] : directions) {
                int newX = i+dx;
                int newY = j+dy;
                if (isValid(newX, newY, n) && grid[newX][newY] != -1) {
                    pq.push({min(maxSafeness, grid[newX][newY]), {newX,newY}});
                }
            }
        }

        return -1;
    }
};
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