<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:
    using state=pair<int,int>;
    int N,M;
    vector<vector<int>>vis;
    vector<vector<int>>dis;
    bool isvalid(int x,int y){
        if(0<=x && x<N && 0<=y && y<M && vis[x][y]==0){
            return 1;
        }
        return 0;
    }
    vector<state>neigh(state next){
        vector<state>neighbour;
        int dx[]={1,-1,0,0};
        int dy[]={0,0,-1,1};
        for(int i=0;i<4;i++){
            int x=dx[i]+next.first;
            int y=dy[i]+next.second;
            if(isvalid(x,y)){
                neighbour.push_back({x,y});
            }
        }
        return neighbour;
    }
    int bfs(vector<state>ones){
        int max_dis=-1;
        queue<state>q;
        for(int i=0;i<ones.size();i++){
            q.push(ones[i]);
        }
        while(!q.empty()){
            state next=q.front();
            q.pop();
            for(state v:neigh(next)){
                if(!vis[v.first][v.second]){
                    vis[v.first][v.second]=1;
                    dis[v.first][v.second]=1+dis[next.first][next.second];
                    q.push(v);
                    max_dis=max(max_dis,dis[v.first][v.second]);
                }
            }
        }
        return max_dis;
    }
    int maxDistance(vector<vector<int>>& grid) {
        N=grid.size();
        M=grid[0].size();
        vector<state>ones;
        vis.assign(N,vector<int>(M,0));
        dis.assign(N,vector<int>(M,0));
        for(int i=0;i<N;i++){
            for(int j=0;j<M;j++){
                if(grid[i][j]==1)
                {
                    ones.push_back({i,j});
                    vis[i][j]=1;
                    dis[i][j]=0;
                }
            }
        }  
        return bfs(ones);

    }
};
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