<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:
    void calculateAllOverSafetyOfCells(vector<vector<int>>& grid, 
    vector<vector<int>>& safety, int n, int xAxis[], int yAxis[]) 
    {
        queue<vector<int>>q ; // steps, row, col
        for(int i=0 ; i<n ; i++)
        {
            for(int j=0 ; j<n ; j++) 
            {
                if(grid[i][j] == 1) 
                {
                    safety[i][j] = 0 ;
                    q.push({0, i, j}) ;
                }
            }
        }

        while(!q.empty())
        {
            auto it = q.front() ;
            q.pop() ;

            int safe = it[0], row = it[1], col = it[2] ;

            for(int idx=0 ; idx<4 ; idx++)
            {
                int dr = row+xAxis[idx] ;
                int dc = col+yAxis[idx] ;

                if(dr>=0 && dr<n && dc>=0 && dc<n && safety[dr][dc] > safe+1)
                {
                    safety[dr][dc] = safe+1 ;
                    q.push({safety[dr][dc], dr, dc}) ;
                }
            }
        }
    }

    bool bfs(vector<vector<int>>& grid, 
    vector<vector<int>>& safety, int n, int xAxis[], int yAxis[], int safeFactor) 
    {
        vector<vector<bool>>vis(n, vector<bool>(n, false)) ;
        queue<pair<int,int>>q ;

        q.push({0, 0}) ;
        vis[0][0] = true ;
        if(safety[0][0] < safeFactor) return false ;

        while(!q.empty())
        {   
            auto it = q.front() ;
            q.pop() ;
            int row = it.first, col = it.second ;

            if(row==n-1 && col==n-1) return true ;

            for(int i=0 ; i<4 ; i++)
            {
                int dr = row+xAxis[i] ;
                int dc = col+yAxis[i] ;

                if(dr>=0 && dr<n && dc>=0 && dc<n && !vis[dr][dc] && safety[dr][dc] >= safeFactor)
                {
                    q.push({dr,dc}) ;
                    vis[dr][dc] = true ;
                }
            }
        }
        return false ;
    }

    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size() ;
        if(grid[0][0]==1 || grid[n-1][n-1]==1) return 0;
        
        vector<vector<int>>safety(n, vector<int>(n, INT_MAX)) ;
        
        int xAxis[] = {-1,0,+1,0} ;
        int yAxis[] = {0,+1,0,-1} ;

        calculateAllOverSafetyOfCells(grid, safety, n, xAxis, yAxis) ;

        int low = 0, high = 400 ;
        while(low <= high)
        {
            int mid = low+(high-low)/2 ;

            if(bfs(grid, safety, n, xAxis, yAxis, mid)) {
                low = mid+1 ;
            }
            else {
                high = mid-1 ;
            }
        }
        return high ;
    }
};
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