<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>

//T.C : O(log(n) * n^2)
//S.C : O(n^2)

class Solution {
public: 
        int dx[4] = {-1, 0, 1, 0};
        int dy[4] = {0, -1, 0, 1};
        int n;
private:
    bool check(vector<vector<int>> &distToNearestThief, int sf)
    {
        queue<pair<int,int>> q;
        
        vector<vector<int>> vis(n, vector<int> (n, 0));
        
        //(0,0) to (n-1, n-1);
        //Applying simple BFS
        
        q.push({0,0});
        vis[0][0] = 1;
        
        if(distToNearestThief[0][0] < sf)
            return false;
        
        while(!q.empty())
        {
            auto curr = q.front();
            q.pop();
            
            int x = curr.first;
            int y = curr.second;
            
            if(x == n-1 && y == n-1)
            {
                return true;
            }
            
            for(int i=0; i<4; i++)
            {
                int a = x+dx[i];
                int b = y+dy[i];
                
                if(a>=0 && b>=0 && a<n && b<n && vis[a][b] == 0 && distToNearestThief[a][b] >= sf)
                {
                    vis[a][b] = 1;
                    q.push({a,b});
                }
            }
        }
        return false;
    }
public:
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        n = grid.size();
        
        //Step 1 - Precalculation of distToNearestThief - for each cell
        vector<vector<int>> distToNearestThief(n, vector<int> (n, -1));
        
        queue<pair<int,int>> q;
        vector<vector<int>> vis(n, vector<int> (n, 0));
        
        //push all cells in queue where thieves are present
        // O(n^2)
        for(int i=0; i<n; i++)
        {
            for(int j=0; j<n; j++)
            {
                if(grid[i][j] == 1)
                {
                    q.push({i,j});
                    vis[i][j] = 1;
                    // distToNearestThief[i][j] = 0;   //dist to a thief to itself is 0
                }
            }
        }
        
        
        // int maxiDist = 0;
        //Applying Multi-source BFS and updating Manhatten distance of each cell to its nearest thief
        
        // O(n^2)
        int level = 0;
        while(!q.empty())
        {
            int size = q.size();
            for(int k=0; k<size; k++)
            {
                auto curr = q.front();
                q.pop();

                int x = curr.first;
                int y = curr.second;
                
                distToNearestThief[x][y]= level;

                for(int i=0; i<4; i++)
                {
                    int a = x+dx[i];
                    int b = y+dy[i];

                    if(a>=0 && b>=0 && a<n && b<n && vis[a][b]== 0)
                    {
                         // distToNearestThief[a][b] = distToNearestThief[x][y] + 1;
                        // maxiDist = max(maxiDist, distToNearestThief[a][b]);         //(for hi)
                        vis[a][b] = 1;
                        q.push({a,b});
                    }
                }
            }
            level++;
        }
        
        
        
        //Step 2: Apply Binary Search on Safe Factor (SF)
        int lo = 0, hi = 400;
        int res = 0;
        
        
        // O(log(n) * n^2)
        while(lo <= hi)
        {
            int mid_sf = lo + (hi - lo)/2;
            //to find a path where every cell should be >= mid
            
            if(check(distToNearestThief, mid_sf))   //if there is a path where all cell >= mid_sf
            {
                res = mid_sf;
                lo = mid_sf + 1;
            }
            else{
                hi = mid_sf - 1;
            }
        }
        
        return res;
    }
};
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