<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>

#include <vector>
#include <queue>
#include <tuple>
#include <algorithm>
using namespace std;

class Solution {
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int nRows = grid.size();
        int nCols = grid[0].size();
        vector<pair<int, int>> directions = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};

        queue<pair<int, int>> thievesGrids;
        for (int row = 0; row < nRows; row++) {
            for (int col = 0; col < nCols; col++) {
                if (grid[row][col] == 1) {
                    thievesGrids.emplace(row, col);
                    grid[row][col] = 0;  // Mark thief cell with 0
                } else {
                    grid[row][col] = -1; // Mark empty cell with -1
                }
            }
        }

        bfs(grid, thievesGrids, nRows, nCols, directions);
        return calculateSafeness(grid, nRows, nCols, directions);
    }

private:
    void bfs(vector<vector<int>>& grid, queue<pair<int, int>>& thievesGrids, int nRows, int nCols, const vector<pair<int, int>>& directions) {
        while (!thievesGrids.empty()) {
            auto [row, col] = thievesGrids.front();
            thievesGrids.pop();
            int value = grid[row][col];

            for (const auto& direction : directions) {
                int newRow = row + direction.first;
                int newCol = col + direction.second;
                if (inBound(newRow, newCol, nRows, nCols) && grid[newRow][newCol] == -1) {
                    grid[newRow][newCol] = value + 1;
                    thievesGrids.emplace(newRow, newCol);
                }
            }
        }
    }

    int calculateSafeness(vector<vector<int>>& grid, int nRows, int nCols, const vector<pair<int, int>>& directions) {
        priority_queue<tuple<int, int, int>> priorityQueue;
        priorityQueue.emplace(grid[0][0], 0, 0);
        vector<vector<bool>> visited(nRows, vector<bool>(nCols, false));
        visited[0][0] = true;

        while (!priorityQueue.empty()) {
            auto [safeness, row, col] = priorityQueue.top();
            priorityQueue.pop();

            if (row == nRows - 1 && col == nCols - 1) {
                return safeness;
            }

            for (const auto& direction : directions) {
                int newRow = row + direction.first;
                int newCol = col + direction.second;
                if (inBound(newRow, newCol, nRows, nCols) && !visited[newRow][newCol]) {
                    visited[newRow][newCol] = true;
                    priorityQueue.emplace(min(safeness, grid[newRow][newCol]), newRow, newCol);
                }
            }
        }

        return -1;
    }

    bool inBound(int row, int col, int nRows, int nCols) {
        return 0 <= row && row < nRows && 0 <= col && col < nCols;
    }
};
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