<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 m, n;
    int dx[4] = {-1, 0, 0, 1};
    int dy[4] = {0, 1, -1, 0};
    
    bool valid(int i, int j) {
        return (i>=0 && j>=0 && i<n && j<n);
    }

    bool ok(vector<vector<int>> dist, int d) {
        queue<pair<int, int>> q;
        vector<vector<bool>> vis(n, vector<bool> (n, false));

        if(dist[0][0] < d)
            return false;

        q.push({0, 0});
        vis[0][0] = true;

        while(!q.empty()) {
            int cx = q.front().first;
            int cy = q.front().second;
            q.pop();

            if(cx == n-1 && cy == n-1)
                return true;

            for(int i=0;i<4;i++) {
                int ni = cx + dx[i];
                int nj = cy + dy[i];

                if(valid(ni, nj) && dist[ni][nj] >= d && !vis[ni][nj]) {
                    q.push({ni, nj});
                    vis[ni][nj] = true;
                }
            }
        }
        return false;
    }

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

        queue<pair<int, int>> q;
        vector<vector<int>> distThief(n, vector<int> (n, -1));
        vector<vector<bool>> vis(n, vector<bool> (n, false));

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

        int lvl=0;
        while(!q.empty()) {
            int sz = q.size();

            while(sz--) {
                int ci = q.front().first;
                int cj = q.front().second;
                q.pop();

                distThief[ci][cj] = lvl;

                for(int i=0;i<4;i++) {
                    int ni = ci + dx[i];
                    int nj = cj + dy[i];

                    if(valid(ni, nj) && !vis[ni][nj]) {
                        q.push({ni, nj});
                        vis[ni][nj] = true;
                    }
                }
            }
            lvl++;
        }
 
        int l=0, r=404;
        int sf=0;

        while(l<=r) {
            int mid = (l+r)/2;

            if(ok(distThief, mid)) {
                sf = mid;
                l = mid+1;
            }
            else 
                r = mid-1;
        }

        return sf;
    }
};
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