<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) {

        int n = grid.size();

        queue<pair<int, int>> multiSourceQueue;

        // Mark thieves as 0 and empty cells as -1, and push thieves to the

        // queue

        for (int i = 0; i < n; i++) {

            for (int j = 0; j < n; j++) {

                if (grid[i][j] == 1) {

                    // Push thief coordinates to the queue

                    multiSourceQueue.push({i, j});

                    // Mark thief cell with 0

                    grid[i][j] = 0;

                } else {

                    // Mark empty cell with -1

                    grid[i][j] = -1;

                }

            }

        }

        // Calculate safeness factor for each cell using BFS

        while (!multiSourceQueue.empty()) {

            int size = multiSourceQueue.size();

            while (size-- > 0) {

                auto curr = multiSourceQueue.front();

                multiSourceQueue.pop();

                // Check neighboring cells

                for (auto& d : dir) {

                    int di = curr.first + d[0];

                    int dj = curr.second + d[1];

                    int val = grid[curr.first][curr.second];

                    // Check if the cell is valid and unvisited

                    if (isValidCell(grid, di, dj) && grid[di][dj] == -1) {

                        // Update safeness factor and push to the queue

                        grid[di][dj] = val + 1;

                        multiSourceQueue.push({di, dj});

                    }

                }

            }

        }

        // Binary search for maximum safeness factor

        int start = 0;

        int end = 0;

        int res = -1;

        for (int i = 0; i < n; i++) {

            for (int j = 0; j < n; j++) {

                // Set end as the maximum safeness factor possible

                end = max(end, grid[i][j]);

            }

        }

        while (start <= end) {

            int mid = start + (end - start) / 2;

            if (isValidSafeness(grid, mid)) {

                // Store valid safeness and search for larger ones

                res = mid;

                start = mid + 1;

            } else {

                end = mid - 1;

            }

        }

        return res;

    }

private:

    // Directions for moving to neighboring cells: right, left, down, up

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

    // Check if a given cell lies within the grid

    bool isValidCell(vector<vector<int>>& grid, int i, int j) {

        int n = grid.size();

        return i >= 0 && j >= 0 && i < n && j < n;

    }

    // Check if a path exists with given minimum safeness value

    bool isValidSafeness(vector<vector<int>>& grid, int minSafeness) {

        int n = grid.size();

        // Check if the source and destination cells satisfy minimum safeness

        if (grid[0][0] < minSafeness || grid[n - 1][n - 1] < minSafeness) {

            return false;

        }

        queue<pair<int, int>> traversalQueue;

        vector<vector<bool>> visited(n, vector<bool>(n, false));

        traversalQueue.push({0, 0});

        visited[0][0] = true;

        // Breadth-first search to find a valid path

        while (!traversalQueue.empty()) {

            auto curr = traversalQueue.front();

            traversalQueue.pop();

            if (curr.first == n - 1 && curr.second == n - 1) {

                return true; // Valid path found

            }

            // Check neighboring cells

            for (auto& d : dir) {

                int di = curr.first + d[0];

                int dj = curr.second + d[1];

                // Check if the neighboring cell is valid and unvisited

                if (isValidCell(grid, di, dj) && !visited[di][dj] &&

                    grid[di][dj] >= minSafeness) {

                    visited[di][dj] = true;

                    traversalQueue.push({di, dj});

                }

            }

        }

        return false; // No valid path found

    }

};

   
        
    

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