<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:
    using pp = pair<int, int>;

    pp M[4] = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

    bool f(int t, vector<vector<int>>& d, vector<vector<int>>& v){
        if(d[0][0] < t)
            return 0;
            
        queue<pp> q;
        q.push({0, 0});
        v[0][0] = 1;

        int n = (int)v.size();

        while(!q.empty()){
            pp x = q.front();
            q.pop();
            for(auto k:M){
                int x1 = x.first + k.first;
                int y1 = x.second + k.second;

                if(x1>=0 and x1<n and y1>=0 and y1<n){
                    if(d[x1][y1] >= t and v[x1][y1]==0){
                        q.push({x1, y1});
                        v[x1][y1] = 1;
                    }
                }
            }
        }
        return v[n-1][n-1]==1;
    }


    int maximumSafenessFactor(vector<vector<int>>& g) {
        int n = (int)g.size(), i, j;

        vector<vector<int>> d(n, vector<int> (n, INT_MAX));
        vector<vector<int>> v(n, vector<int> (n, 0));


        queue<pp> q;

        for(i=0;i<n;i++){
            for(j=0;j<n;j++){
                if(g[i][j]){
                    q.push({i, j});
                    d[i][j] = 0;
                }
            }
        }

        while(!q.empty()){
            pp t = q.front();
            q.pop();

            for(auto i:M){
                int x = t.first + i.first, y = t.second + i.second;
                if(x>=0 and x<n and y>=0 and y<n){
                    if(d[x][y]==INT_MAX){
                        d[x][y] = 1 + d[t.first][t.second];
                        q.push({x, y});
                    }
                }
            }
        }

        i = 0;
        j = n;
        int ans = 0;

        while(i<=j){
            fill(v.begin(), v.end(), vector<int> (n, 0));
            int m = i + (j-i)/2;
            if(f(m, d, v)){
                ans = m;
                i = m + 1;
            }else
                j = m - 1;
        }
        
        return ans;
       
    }
};
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