<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>

#define pi pair<int,int>
#define mk make_pair
class Solution {
public:

    bool valid(int i,int j,int n){
        if (i<0 or j<0 or i>=n or j>=n) return false;
        return true;
    }
    int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        queue<pi> q;
        map<pi,int> dis;
         map<pi,bool> vis;
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                // cout<<23<<endl;
                if (grid[i][j]==1){
                    dis[mk(i,j)] = 0;
                    vis[mk(i,j)] = true;

                   q.push(mk(i,j));
                }
            }
        }
       
        int ans = -1;
        vector<pi> pos = {{0,1},{1,0},{0,-1},{-1,0}};
        while(!q.empty()){
            pi p  = q.front();
            q.pop();
            for (auto &i: pos){
                int nx = i.first+p.first;
                int ny = i.second+p.second;
                if (valid(nx,ny,n) and !vis[mk(nx,ny)]) {
                    vis[(mk(nx,ny))]  = true;
                    dis[mk(nx,ny)] = 1 + dis[p];
                    q.push(mk(nx,ny));
                }
            }
        }
       
        for (int i=0;i<n;i++){
            for (int j=0;j<n;j++){
                // cout<<23<<endl;
                if (grid[i][j]==0){
                    pi p = mk(i,j);
                    if (dis.find(p)!=dis.end()){
                        ans = max(ans,dis[p]);
                    }
                }
            }
        }
        return ans;
        
    }
};
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