<p>Given an <code>n x n</code> <code>grid</code>&nbsp;containing only values <code>0</code> and <code>1</code>, where&nbsp;<code>0</code> represents water&nbsp;and <code>1</code> represents land, find a water cell such that its distance to the nearest land cell is maximized, and return the distance.&nbsp;If no land or water exists in the grid, return <code>-1</code>.</p>

<p>The distance used in this problem is the Manhattan distance:&nbsp;the distance between two cells <code>(x0, y0)</code> and <code>(x1, y1)</code> is <code>|x0 - x1| + |y0 - y1|</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex1.JPG" style="width: 185px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,1],[0,0,0],[1,0,1]]
<strong>Output:</strong> 2
<strong>Explanation:</strong> The cell (1, 1) is as far as possible from all the land with distance 2.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2019/05/03/1336_ex2.JPG" style="width: 184px; height: 87px;" />
<pre>
<strong>Input:</strong> grid = [[1,0,0],[0,0,0],[0,0,0]]
<strong>Output:</strong> 4
<strong>Explanation:</strong> The cell (2, 2) is as far as possible from all the land with distance 4.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>n == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= n&nbsp;&lt;= 100</code></li>
	<li><code>grid[i][j]</code>&nbsp;is <code>0</code> or <code>1</code></li>
</ul>

class Solution {
public:
    int maxDistance(vector<vector<int>>& grid) {
      int m=grid.size();
      int n=grid[0].size();
      vector<vector<bool>> vis(m,vector<bool>(n,false));
      queue<pair<int,int>> q;
      for(int i=0;i<m;i++){
        for(int j=0;j<n;j++){
            if(grid[i][j]==1){
                q.push({i,j});
            }
        }
      }
      vector<pair<int,int>> dir={{-1,0},{0,-1},{0,1},{1,0}};
      if(q.empty() || q.size()==m*n) return -1;
      int level=0;
      while(!q.empty()) {
          int n1=q.size();
          for(int i=0;i<n1;i++){
            int row=q.front().first;
            int col=q.front().second;
            q.pop();
              for(auto d:dir){
                int nr=row+d.first;
                int nc=col+d.second;
                if(nr>=0 && nr<m && nc>=0 && nc<n && grid[nr][nc]==0){
                    q.push({nr,nc});
                    grid[nr][nc]=-1;
                }
              }
          }
          level++;
      }
      return level-1;
    }
};
23181