<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 ppi pair<int,int>
class Solution {
public:
    int n;
    queue<ppi> q;
    vector<ppi> dir={{0,1} ,{0,-1} ,{1,0} ,{-1,0}};
    unordered_set<string> visited;
    string trr(int i,int j)
    {
        return to_string(i)+"."+to_string(j);
    }
    int bfs(vector<vector<int>>& grid)
    {
        int max_d=-1;
        //initialize hho chhuka hh
        //seedha bfs lagao
        while(!q.empty())
        {
            // cout<<"yy";
            int s=q.size();
            for(int it=0;it<s;it++)
            {
                //front nikaalo
                int i=q.front().first;
                int j=q.front().second;
                q.pop();
                //neighbour check
                for(auto x:dir)
                {
                    int r= i+x.first;
                    int c= j+x.second;
                    if(r<0 || c<0 || r>=n || c>=n) continue;
                    if(grid[r][c]==0 && visited.find(trr(r,c)) == visited.end())
                    {
                        cout<<r<<" "<<c<<endl;
                        q.push({r,c});
                        visited.insert(trr(r,c));
                    }
                }
            }
            max_d++;
        }
        return max_d;
    }
    int maxDistance(vector<vector<int>>& grid) 
    {
        n=grid.size();
        int zeroes=0,ones=0;
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<n;j++)
            {
                if(grid[i][j]==1) {q.push({i,j});ones++;}
                else if(grid[i][j]==0) zeroes++;
            }
        }
        if(ones==n*n || zeroes==n*n) return -1;
        return bfs(grid);
    }
};
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