<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<pair<int, int>, int> > q;
        int n=grid.size();
        int m=grid[0].size();

        vector<vector<int> > vis(n, vector<int>(m, 0));

        vector<vector<int> > dis(n, vector<int>(m, -1));

        cout<<"here1\n";

        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[i].size();j++){
                if(grid[i][j]==1){
                    q.push(make_pair(make_pair(i, j), 0));
                    dis[i][j]=0;
                    vis[i][j]=1;
                }
            }
        }

        cout<<"here2\n";

        while(!q.empty()) {
            pair<pair<int, int>, int> front = q.front();

            cout<<"here3\n";

            int x=front.first.first;
            int y=front.first.second;
            int dist=front.second;
            cout<<"x: "<<x<<" ,y: "<<y<<" ,dist: "<<dist<<endl;
            q.pop();

            dis[x][y]=max(dis[x][y], dist);

            vector<int> dx = {-1, 1, 0, 0};
            vector<int> dy = {0, 0, -1, 1};

            for(int i=0;i<4;i++){
                int nx=x+dx[i];
                int ny=y+dy[i];

                if(nx>=n || nx<0 || ny>=m || ny<0)continue;

                if(vis[nx][ny]==1)continue;

                vis[nx][ny]=1;

                int ndis=abs(nx-x)+abs(ny-y)+dist;
                q.push(make_pair(make_pair(nx, ny), ndis));
            }



        }

        int ans=-1;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j]==0)
                ans=max(ans, dis[i][j]);
            }
        }

        return ans;




        
    }
};
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