<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 = grid.size();
        int m = grid[0].size();
        if (grid[0][0] == 1 || grid[n - 1][m - 1] == 1) {
            return 0;
        }

        vector<vector<int>> v(n, vector<int>(m, 0));
        vector<vector<int>> distance(n, vector<int>(m, INT_MIN));
        vector<vector<bool>> visited(n, vector<bool>(m, false));
        queue<pair<int, int>> q;
        int ans = INT_MAX;

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

        while (!q.empty()) {
            pair<int, int> p = q.front();
            q.pop();
            int r = p.first;
            int c = p.second;

            int move_x[] = {-1, 1, 0, 0};
            int move_y[] = {0, 0, -1, 1};

            for (int i = 0; i < 4; i++) {
                int row = r + move_x[i];
                int col = c + move_y[i];

                if (row >= 0 && row < n && col >= 0 && col < m &&
                    !visited[row][col]) {
                    grid[row][col] = grid[r][c] + 1;
                    q.push({row, col});
                    visited[row][col] = true;
                }
            }
        }

        //  for (int i = 0; i < n; i++) {
        //     for (int j = 0; j < m; j++) {
        //        cout<<grid[i][j]<<" ";
        //     }
        //     cout<<"\n";
        // }

        deque<pair<int, pair<int, int>>> pq;
        vector<vector<bool>> visited1(n, vector<bool>(m, false));

        pq.push_front({grid[0][0], {0, 0}});
        distance[0][0] = grid[0][0];
        visited1[0][0] = true;

        while (!pq.empty()) {
            pair<int, pair<int, int>> p = pq.front();
            pq.pop_front();
            int dis = p.first;
            int row = p.second.first;
            int col = p.second.second;

            //cout<<"grid[row][col]: "<<grid[row][col]<<"\n";

            ans = min(ans, grid[row][col]);
            
            if(row == n-1 && col == m-1){
                //cout<<dis<<"\n";
                //return dis;
                break;
            }

            // if (visited1[row][col]) {
            //     continue;
            // }

           
            //cout<<grid[row][col]<<" ";
            
           

            int move_x[] = {-1, 1, 0, 0};
            int move_y[] = {0, 0, 1, -1};

            for (int i = 0; i < 4; i++) {
                int r = row + move_x[i];
                int c = col + move_y[i];

                if (r >= 0 && r < n && c >= 0 && c < m &&
                    !visited1[r][c]) {
                    //distance[r][c] = dis + grid[r][c];
                    if(grid[r][c] >= ans){
                        pq.push_front({grid[r][c], {r,c}});
                    }
                    else{
                        pq.push_back({dis, {r,c}});
                    }
                    //pq.push({min(grid[r][c],dis), {r, c}});
                    visited1[r][c] = true;
                }
            }
        }

        //cout<<"\n"<<distance[n-1][m-1];
        //cout<<"ans is: "<<ans<<"\n";
        return ans;
    }
};
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