<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:
    vector<pair<int, int>> move = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};

    bool isValidSafeness(vector<vector<int>>& grid, int minSafeness) {
        int n = grid.size();

        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;

        while (!traversalQueue.empty()) {
            auto curr = traversalQueue.front();
            traversalQueue.pop();
            if (curr.first == n - 1 && curr.second == n - 1) {
                return true;
            }
            for (int i = 0; i < 4; i++) {
                int di = curr.first + move[i].first;
                int dj = curr.second + move[i].second;
                if (di >= 0 && dj >= 0 && di < n && dj < n &&
                    !visited[di][dj] && grid[di][dj] >= minSafeness) {
                    visited[di][dj] = true;
                    traversalQueue.push({di, dj});
                }
            }
        }
        return false;
    }

    int maximumSafenessFactor(vector<vector<int>>& grid) {

        queue<pair<int, int>> q;
        int n = grid.size();
        // vector<vector<int>> dis(n, vector<int>(n, INT_MAX));

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

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j]) {
                    q.push({i, j});
                    grid[i][j] = 0;
                } else
                    grid[i][j] = -1;
            }
        }

        while (q.empty() == false) {

            int size = q.size();
            while (size-- > 0) {
                auto curr = q.front();
                q.pop();
                // Check neighboring cells
                 for (int i = 0; i < 4; i++) {
                    int di = curr.first + move[i].first;
                    int dj = curr.second + move[i].second;
                    int val = grid[curr.first][curr.second];

                    if (di >= 0 && dj >= 0 && di < n && dj < n && grid[di][dj] == -1) {
                        grid[di][dj] = val + 1;
                        q.push({di, dj});
                    }
                }
            }
        }

        int start = 0;
        int end = 0;
        int res = -1;

        for (int i = 0; i < n; i++)
            for (int j = 0; j < n; j++) 
                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;
    }
};
156948