<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 i, j, m = grid.size(), n = grid[0].size();
        queue<vector<int>> q;
        for (i = 0; i < m; i++)
        {
            for (j = 0; j < n; j++)
            {
                if (grid[i][j])
                {
                    q.push({i, j});
                    grid[i][j] = 2;
                }
            }
        }
        int x = 0;
        vector<int> v;
        while (!q.empty())
        {
            int n1 = q.size();
            x++;
            while (n1--)
            {
                v = q.front();
                // cout << v[0] << " " << v[1] << endl;
                q.pop();
                if (v[0] - 1 >= 0 && grid[v[0] - 1][v[1]] == 0)
                {
                // cout<<v[0]-1<<" "<<v[1]<<" #"<<endl;
                    q.push({v[0] - 1, v[1]});
                    grid[v[0] - 1][v[1]] = 2;
                }
                if (v[0] + 1 < m && grid[v[0] + 1][v[1]] == 0)
                {
                // cout<<v[0]+1<<" "<<v[1]<<" #"<<endl;
                    q.push({v[0] + 1, v[1]});
                    grid[v[0] + 1][v[1]] = 2;
                }
                if (v[1] - 1 >= 0 && grid[v[0]][v[1] - 1] == 0)
                {
                // cout<<v[0]<<" "<<v[1]-1<<" #"<<endl;
                    q.push({v[0], v[1] - 1});
                    grid[v[0]][v[1] - 1] = 2;
                }
                if (v[1] + 1 < n && grid[v[0]][v[1] + 1] == 0)
                {
                // cout<<v[0]<<" "<<v[1]+1<<" #"<<endl;
                    q.push({v[0], v[1] + 1});
                    grid[v[0]][v[1] + 1] = 2;
                }
            }

            // cout << endl;
        }
        return x - 1==0?-1:x-1;
    }
};
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