<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){

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

        for(int i = 0; i < n; i++)
        {
            for(int j=0; j < m; j++)
            {
                if(grid[i][j] == 1) q.push({i,j});
            }
        }
        
        int  res = 0;

        while(!q.empty())
        {
            
            int temp_i = q.front().first;
            int temp_j = q.front().second;
            q.pop();
          
            vector<pair<int,int>>dir = {{0,1},{0,-1},{1,0},{-1,0}};
           
            for(auto &it : dir)
            {
               
                int new_i= it.first + temp_i;
                int new_j= it.second + temp_j;
               
                if(new_i>=0 && new_i<n && new_j>=0 && new_j<m && grid[new_i][new_j]==0)
                {
                    
                    q.push({new_i,new_j});
                   
                    grid[new_i][new_j]=grid[temp_i][temp_j]+1;
                    
                    res = max(res, grid[new_i][new_j]);
                }
            }
        }
    
        return res - 1;
    }
};
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