<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>> visited;
    vector<vector<int>> dist;
    vector<int> dx;
    vector<int> dy;
    bool check(int x,int y,vector<vector<int>>& grid){
        if(x<0 || x>=grid.size() || y<0 || y>=grid.size()) return false;
        return true;
    }
    vector<pair<int,int>> neighbour(pair<int,int>node ,vector<vector<int>>& grid ){
        vector<pair<int,int>> ans;
        for(int i=0;i<4;i++){
            int newx = node.first + dx[i];
            int newy = node.second + dy[i];
            if(check(newx,newy,grid)){
                ans.push_back({newx,newy});
            }
        }
        return ans;
    }
    void bfs(vector<vector<int>>& grid){
        queue<pair<int,int>> q;
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[0].size();j++){
                if(grid[i][j] == 1){
                    q.push({i,j});
                    dist[i][j]=0;
                }
            }
        }
        while(!q.empty()){
            pair<int,int> tp=q.front();
            q.pop();
            if(visited[tp.first][tp.second]) continue;
            visited[tp.first][tp.second]=1;
            for(auto &v: neighbour(tp,grid)){
                if(dist[v.first][v.second] > dist[tp.first][tp.second] +1 ){
                    q.push(v);
                    dist[v.first][v.second] = dist[tp.first][tp.second] +1 ;
                }
            }
        }
    }
    int maxDistance(vector<vector<int>>& grid) {
        dy = {1,0,-1,0};
        dx = {0,1,0,-1};
        int n=grid.size();
        int m=grid[0].size();
        visited.assign(n,vector<int>(m,0));
        dist.assign(n,vector<int>(m,1e9));
        bfs(grid);
        int ans=0;
        for(int i=0;i<grid.size();i++){
            for(int j=0;j<grid[0].size();j++){
                ans=max(dist[i][j],ans);
            }
        }
        return (ans==0 || ans==1e9)?-1:ans;
    }
};
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