<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>> arr = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
    bool dfs(int n, int i, int j, int mid, vector<vector<int>>& dist, vector<vector<int>>& visited) {
        if (dist[i][j] < mid) {
            return false;
        }
        if (i == n - 1 && j == n - 1) {
            return true;
        }
        visited[i][j] = 1;
        for (int k = 0; k < 4; k++) {
            int ni = i + arr[k].first;
            int nj = j + arr[k].second;
            if (ni >= 0 && ni < n && nj >= 0 && nj < n && !visited[ni][nj]) {
                if (dfs(n, ni, nj, mid, dist, visited)) {
                    return true;
                }
            }
        }
        return false;
    }
    int result(int step, vector<vector<int>>& dist) {
        int ans = 0;
        int l = 0;
        int r = step;
        while (l <= r) {
            int mid = (l + r) / 2;
            vector<vector<int>> visited(dist.size(), vector<int>(dist[0].size(), false));
            if (dfs(dist.size(), 0, 0, mid, dist, visited)) {
                ans = mid;
                l = mid + 1;
            }
            else {
                r = mid - 1;
            }
        }
        return ans;
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size(), m = grid[0].size();
        queue<pair<int,int>> q;
        vector<vector<int>> dist(n,vector<int> (m,0));
        vector<vector<int>> vis(n,vector<int> (m,0));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++){
                if (grid[i][j] == 1) {
                    q.push({i, j});
                    vis[i][j] = 1;
                }    
            }
        }    
        int step = 0;
        while (!q.empty()) {
            int size = q.size();
            while (size--) {
                auto front = q.front();
                int r = front.first;
                int c = front.second;
                q.pop();
                vector<pair<int, int>> arr = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
                for (auto& dir : arr) {
                    int ni = r + dir.first;
                    int nj = c + dir.second;
                    if (ni >= 0 && ni < grid.size() && nj >= 0 && nj < grid[0].size() 
                    && !vis[ni][nj]) {
                        q.push({ni, nj});
                        vis[ni][nj] = 1;
                        dist[ni][nj] = step + 1;
                    }
                }
            }
            step++;
        }
        return result(step, dist);
    }
};
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