<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:
vector<vector<int>> dirs = {{0,1},{1,0},{0,-1},{-1,0}};
    int maxDistance(vector<vector<int>>& grid) {
        int m=grid.size(),n = grid[0].size();
        vector<vector<bool>> visited(m,vector<bool>(n,false));
        int res = -1;
        queue<vector<int>> q;
        for(int i =0; i<m ;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    visited[i][j]=true;
                    q.push({i,j,0});
                }
            }
        }
        while(!q.empty()){
            vector<int> t = q.front();
            q.pop();
            for(auto & dir : dirs){
                int x= t[0]+dir[0];
                int y = t[1]+ dir[1];
                if ( x<0 || x>= m || y<0 || y>=n) continue;
                if( visited[x][y]) continue;
                q.push({x,y,t[2]+1});
                visited[x][y]=true;
                res=max(res,t[2]+1);
            }
        }return res;
    }
};
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