<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 {
    vector<pair<int, int>> dirs = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    bool isValid(int i, int j, vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        if (i >= 0 && i < n && j >= 0 && j < m)
            return true;

        return false;
    }

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

        if (grid[0][0] || grid[n - 1][m - 1])
            return 0;

        queue<vector<int>> q;

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

        while (!q.empty()) {
            vector<int> top = q.front();
            q.pop();
            int i = top[0];
            int j = top[1];
            int d = top[2];

            for (auto dir : dirs) {
                int x = i + dir.first;
                int y = j + dir.second;

                if (!isValid(x, y, grid) || grid[x][y] != INT_MAX) {
                    continue;
                }
                grid[x][y] = d + 1;
                q.push({x, y, d + 1});
            }
        }

         priority_queue<vector<int>> pq;
        // Push starting cell to the priority queue
        pq.push(vector<int>{grid[0][0], 0, 0}); // [maximum_safeness_till_now, x-coordinate, y-coordinate]
        grid[0][0] = -1; // Mark the source cell as visited

        // BFS to find the path with maximum safeness factor
        while (!pq.empty()) {
            auto curr = pq.top();
            pq.pop();

            // If reached the destination, return safeness factor
            if (curr[1] == n - 1 && curr[2] == n - 1) {
                return curr[0];
            }

            // Explore neighboring cells
            for (auto& d : dirs) {
                int di = d.first + curr[1];
                int dj = d.second + curr[2];
                if (isValid(di, dj,grid) && grid[di][dj] != -1) {
                    // Update safeness factor for the path and mark the cell as visited
                    pq.push(vector<int>{min(curr[0], grid[di][dj]), di, dj});
                    grid[di][dj] = -1;
                }
            }
        }

        return -1;







       

        for (int i = 1; i < n; i++) {
            grid[i][0] = min(grid[i][0], grid[i - 1][0]);
            grid[0][i] = min(grid[0][i], grid[0][i - 1]);
        }

        for (int i = 1; i < n; i++) {
            for (int j = 1; j < m; j++) {
                grid[i][j] =
                    min(grid[i][j], max(grid[i - 1][j], grid[i][j - 1]));
            }
           // cout << "" << endl;
        }
        return grid[n - 1][m - 1];
    }
};
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