<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:
/*
 Nice: think in the reverse direction i.e. in order to find the shortest 
 distance  for each cell to to theives (1) whcih will result in tc (n^2*m^2) , 
think in the opposite fashion i.e. try to start from  1s and reach every 0s in
minimum possible path using multi source BFS in (n*m) time

algo:
1. precompute the shortest distance of each cell(0) to the nearest theif
2. do binary search on answer i.e disatance [0...800]
3. check function: is it possible to reach d from s with mid as the safness factor
   condition: shortest dist val at each cell>=  safeness factor else it will act as the blockage
4. print the maximum safeness factor ans the answer

*/
    vector<int>dr={0,0,-1,1}, dc= {1,-1,0,0};  
    int calMinShortestDist(vector<vector<int>>&grid, vector<vector<int>>&shortestPath){
        int n= grid.size();
        queue< pair<int, int>>q;// dist,{r,c}
        vector<vector<int>>vis(n, vector<int>(n,0));
        // push all the thieves into the queue
        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;
                }
            }
        }
        int len=0;
// Multi sources BFS
        while(!q.empty()){
            int size = q.size();
            while(size--){
               auto curr = q.front();
               q.pop();           
               int i= curr.first;
               int j= curr.second;
               shortestPath[i][j]= len;
               
               for(int d=0; d<4; d++){
                 int ni = i + dr[d], nj = j + dc[d];
                 if(ni>=0 && ni<n && nj>=0 && nj<n && vis[ni][nj]==0) {
                        q.push({ni,nj});
                        vis[ni][nj]=1;
                 }
               }
            }
            len++;
        }  
        return len;
        
    }


     bool isPossible(int safenessFactor, vector<vector<int>>&grid, vector<vector<int>>& shortestPath) {
        if (shortestPath[0][0] < safenessFactor) return false;
        int n= grid.size();
        queue<pair<int, int>> q;
        q.push({0, 0});
        vector<vector<int>> vis(n, vector<int>(n, 0));
        vis[0][0] = 1;

        while (!q.empty()) {
            auto [i, j] = q.front();
            q.pop();

            if (i == n - 1 && j == n - 1) return true;

            for (int d = 0; d < 4; d++) {
                int ni = i + dr[d], nj = j + dc[d];
                if (ni >= 0 && ni < n && nj >= 0 && nj < n && !vis[ni][nj] && shortestPath[ni][nj] >= safenessFactor) {
                    vis[ni][nj] = 1;
                    q.push({ni, nj});
                }
            }
        }

        return false;
}

    int maximumSafenessFactor(vector<vector<int>>& grid) {
         int n = grid.size();
         vector<vector<int>>shortestPath(n, vector<int>(n,1e5));
         int len = calMinShortestDist(grid,shortestPath);
         for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                cout<<shortestPath[i][j]<<" ";
            }cout<<"\n";
         }
        int l=0, hi= len, ans=0;
        while(l<=hi){
            int mid = l + (hi-l)/2;           
            if(isPossible(mid,grid,shortestPath)) {
                ans = mid;
                l= mid +1;
            }
            else hi= mid-1;
        }
        // if(isPossible(0,grid,shortestPath)) cout<<"yes";
        // if(isPossible(3,grid,shortestPath)) cout<<"yes";
        // if(isPossible(0,grid,shortestPath)) cout<<"yes";
        // else cout<<"no";
        return ans;
    }
};
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