<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>> dis;
    vector<vector<int>> vis;

    void bfs(queue<pair<int,int>> q,vector<vector<int>>& grid)
    {
        int n=grid.size();
        while(!q.empty())
        {
            auto node=q.front();
            q.pop();
            if(vis[node.first][node.second]==1)
            {
                continue;
            }
            vis[node.first][node.second]=1;
            int dx[]={-1,0,1,0};
            int dy[]={0,1,0,-1};
            for(int i=0;i<4;i++)
            {
                int x=node.first+dx[i];
                int y=node.second+dy[i];
                if(x>=0 && x<n && y>=0 && y<n && vis[x][y]!=1 && grid[x][y]==0)
                {
                    if(dis[x][y]>dis[node.first][node.second]+1)
                    {
                        dis[x][y]=dis[node.first][node.second]+1;
                        q.push({x,y});
                    }
                }
            }
        }
    }


    int maxDistance(vector<vector<int>>& grid) {
        int n=grid.size();
        for(int i=0;i<n;i++)
        {
            vector<int> v(n,0);
            vector<int> d(n,1e9);
            vis.push_back(v);
            dis.push_back(d);
        }
        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;
                    //vis[i][j]=1;
                }
            }
        }
        //cout<<q.size()<<" "<<n*n<<endl;
        if(q.size()==n*n || q.size()==0)
        {
            return -1;
        }
        bfs(q,grid);
        int ans=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                ans=max(ans,dis[i][j]);
            }
        }
        return ans;
    }
};
30531