<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 buildDis(queue<pair<int,int>>& q, int n, vector<vector<int>>& dis, vector<vector<bool>>& vis){
    //     vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    //     int lvl = 0;

    //     while (!q.empty()) {
    //         int siz = q.size();
    //         while(siz--){
    //             auto [x,y] = q.front();
    //             q.pop();
    //             dis[x][y] = lvl;
    //             for (int i=0; i<4; i++) {
    //                 int nx = x+dir[i].first, ny = y+dir[i].second;
    //                 if (nx>=0 && nx<n && ny>=0 && ny<n && !vis[nx][ny]) {
    //                     q.push({nx,ny});
    //                     vis[nx][ny] = true;
    //                 }
    //             }
    //         }
    //         lvl++;
    //     }
    // }

    // int findSf(vector<vector<int>>& dis, int n){
    //     priority_queue<pair<int,pair<int,int>>> pq;
    //     vector<vector<bool>> vis (n,vector<bool>(n));
    //     pq.push({dis[0][0],{0,0}});
    //     vis[0][0] = true;
    //     vector<pair<int, int>> dir = {{0, 1}, {0, -1}, {1, 0}, {-1, 0}};
    //     int ans = 1e9;
    //     while(!pq.empty()){
    //         auto [d,cord] = pq.top();
    //         pq.pop();
    //         int x = cord.first, y = cord.second;
    //         ans = min(ans,dis[x][y]);
    //         if(x==n-1 && y==n-1) return ans;
    //         for(int i=0; i<4; i++){
    //             int nx = x+dir[i].first, ny = y+dir[i].second;
    //             if(nx>=0 && nx<n && ny>=0 && ny<n &&!vis[nx][ny]){
    //                 vis[nx][ny] = true;
    //                 pq.push({dis[nx][ny],{nx,ny}});
    //             }
    //         }
    //     }
    //     return ans;
    // }

//     int maximumSafenessFactor(vector<vector<int>>& grid) {
//         int n = grid.size();
        
//         queue<pair<int, int>> q;
//         vector<vector<bool>> vis (n,vector<bool>(n));

//         for (int r = 0; r < n; r++) {
//             for (int c = 0; c < n; c++) {
//                 if (grid[r][c] == 1) {
//                     vis[r][c] = true;
//                     q.push({r, c});
//                 }
//             }
//         }
        
//         vector<vector<int>> dis (n,vector<int>(n));

//         buildDis(q,n,dis,vis);
//         int ans = findSf(dis,n);

//         return ans;
//     }
// };

    void buildDis(queue<pair<int,int>>& q, int n, vector<vector<int>>& dis, vector<vector<bool>>& vis){

        int dx[4] = {0,0,1,-1} ;
        int dy[4] = {1,-1,0,0} ;
        int dist = 0 ;

        while( !q.empty() ){
            int size = q.size() ;
            while( size-- ){
                int x = q.front().first ;
                int y = q.front().second ;
                dis[x][y] = dist ;
                q.pop() ;
                for( int i=0 ; i<4 ; i++ ){
                    int new_x = x+dx[i] ;
                    int new_y = y+dy[i] ;
                    if( new_x>=0 && new_y>=0 && new_x < n && new_y < n && !vis[new_x][new_y] ){
                        vis[new_x][new_y] = 1 ;
                        q.push( { new_x , new_y } ) ;
                    }
                }
            }
            dist++ ;
        }
    }

    int findSafest( vector<vector<int>>& dist , int n ){

        vector<vector<bool>> vis( n , vector<bool>( n , 0 ) ) ;

        int dx[4] = {0,0,-1,1} ;
        int dy[4] = {1,-1,0,0} ;

        priority_queue<pair<int,pair<int,int>>> pq ;
        pq.push( { dist[0][0] , {0,0} } ) ;
        vis[0][0] = true ;

        int ans = dist[0][0] ;

        while(!pq.empty()){

            int min_safe_dist = pq.top().first ;
            pair<int,int> curr = pq.top().second ;
            int x = curr.first ; int y = curr.second ;
            pq.pop() ;

            ans = min( ans , min_safe_dist ) ;
            if( curr.first == n-1 && curr.second == n-1 ) break ;

            for( int i=0 ; i<4 ; i++ ){
                int new_x = x+dx[i] ;
                int new_y = y+dy[i] ;
                if( new_x>=0 && new_y>=0 && new_x<n && new_y<n && !vis[new_x][new_y] ){
                    vis[new_x][new_y] = 1 ;
                    pq.push( { min( min_safe_dist , dist[new_x][new_y] ) , { new_x , new_y } } ) ;
                }
            }
        }
        return ans ;
    }

    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size() ;

        queue<pair<int,int>> q ;
        vector<vector<bool>>vis( n , vector<bool> ( n , 0 ) ) ;

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

        vector<vector<int>> dis( n, vector<int>( n , 0 ) ) ;

        buildDis( q , n , dis , vis ) ;
        int ans = findSafest( dis , n ) ;

        return ans;
    }
};

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