<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 = int(grid.size());

        vector<vector<int>> dists(n, vector<int>(n, -1));

        queue<pair<int, int>> q;

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

        while (!q.empty())
        {
            auto t = q.front(); q.pop();
            int x = t.first, y = t.second;
            int dist = dists[x][y];

            if (y - 1 >= 0 && dists[x][y - 1] == -1) dists[x][y - 1] = dist + 1, q.push({ x,y - 1 });
            if (y + 1 < n && dists[x][y + 1] == -1) dists[x][y + 1] = dist + 1, q.push({ x,y + 1 });
            if (x - 1 >= 0 && dists[x - 1][y] == -1) dists[x - 1][y] = dist + 1, q.push({ x - 1,y });
            if (x + 1 < n && dists[x + 1][y] == -1) dists[x + 1][y] = dist + 1, q.push({ x + 1,y });

        }

        int left=0,right = n,r=0;

        while(left<=right)
        {
            int test = (left+right)/2;

             if(feasible(dists,n,test))
             {
                r=test;
                left=test+1;
             }
             else
             right=test-1;
        }

        return r;

    }

    bool feasible(const vector<vector<int>>& dists, int n, int testDist)
    {
         queue<pair<int, int>> q;
         q.push({0,0});

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

        while (!q.empty())
        {
            auto t = q.front(); q.pop();
            int x = t.first, y = t.second;

            if(visited[x][y]) continue;

            int dist = dists[x][y];
              if(dist<testDist) continue;

              if(x==n-1&&y==n-1) return true;

            if (y - 1 >= 0) q.push({ x,y - 1 });
            if (y + 1 < n) q.push({ x,y + 1 });
            if (x - 1 >= 0) q.push({ x - 1,y });
            if (x + 1 < n) q.push({ x + 1,y });

            visited[x][y] = true;

        }

        return false;
    }

};
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