<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 poss(vector<vector<int>> &dis, int val, vector<vector<int>> &vis){
        queue<pair<int,int>>q;
        int n = dis.size();
        if(dis[0][0] < val)return 0;
        vis[0][0] = val + 1;
        q.push({0,0});
        // vis[0][0] = 1;
        int dx[] = {0, 1, 0, -1};
        int dy[] = {1, 0, -1, 0};
        while(!q.empty()){
            auto [x, y] = q.front();q.pop();
            // int x = ele[0], y = ele[1];
            if(x == n - 1 && y == n - 1)return 1;
            for(int i = 0; i < 4; i++){
                int newx = x + dx[i], newy = y + dy[i];
                // if(newx < 0 || newx >=n || newy < 0 || newy >=n || vis[newx][newy] == val + 1)continue;
                // if(dis[newx][newy] < val)continue;
                if(newx >= 0 && newx < n && newy >= 0 && newy < n && vis[newx][newy] != val + 1 && dis[newx][newy] >=val){
                q.push({newx, newy});
                vis[newx][newy] = val + 1;
                }
            }
        }
        return 0;
    }
    int maximumSafenessFactor(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>>dis(n, vector<int>(n, 1e9));
        queue<pair<int,int>>q;
        for(int i = 0; i < n; i++){
            for(int j = 0; j < n; j++){
                if(grid[i][j] == 1){
                    q.push({i,j});
                    dis[i][j] = 0;
                }
            }
        }
        int dx[] = {0, 1, 0, -1};
        int dy[] = {1, 0, -1, 0};
        int len = 1;
        while(!q.empty()){
            int s = q.size();
            for(int i = 0; i < s; i++){
                auto [x, y] = q.front();
                q.pop();
                // int x = a[0], y = a[1];
                for(int j = 0; j < 4; j++){
                    int newx = x + dx[j], newy = y + dy[j];
                    // if(newx < 0 || newx >= n || newy < 0 || newy >=n)continue;
                    // if(dis[newx][newy] <= len)continue;
                    if(newx >= 0 && newx < n && newy >= 0 && newy < n  && dis[newx][newy] > len){
                    q.push({newx, newy});
                    dis[newx][newy] = len;
                    }
                }
            }
            len++;
        }
        int l = 0, r = 400;
        int ans = 0;
        vector<vector<int>>vis(dis.size(), vector<int>(dis.size(), 0));

        while(l <= r){
            int mid = (l + r)/2;
            if(poss(dis, mid, vis)){
                ans = mid;
                l = mid + 1;
            } else r = mid -1;
        }
        return ans;
    }
};
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