<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) {
        queue<pair<int,int>> q;
        map<pair<int,int>,int> mp;
        vector<pair<int,int>> dirs={{0,1},{0,-1},{1,0},{-1,0}};
        for(int x=0;x<grid.size();x++){
            for(int y=0;y<grid[0].size();y++){
                if(grid[x][y]==1){
                    q.push({x,y});
                    mp[{x,y}]=0;
                }
            }
        }
        while(!q.empty()){
            pair<int,int> c=q.front();
            q.pop();
            int cx=c.first, cy=c.second;
            for(auto d:dirs){
                int nx=cx+d.first, ny=cy+d.second;
                if(nx>=0 && nx<grid.size() && ny>=0 && ny<grid[0].size() && grid[nx][ny]==0){
                    if(mp.find({nx,ny})==mp.end()){
                        q.push({nx,ny});
                        mp[{nx,ny}]=mp[{cx,cy}]+1;
                    }
                }
            }
        }
        int a=0;
        for(auto i:mp){
            a=max(a,i.second);
        }
        if(a>0 )return a;
        return -1;
    }
};
44006