<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 ans = INT_MIN;
        queue<vector<int>> q;
        int n = grid.size();

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

        if(q.size()==0 || q.size()==n*n)
        {
            return -1;
        }

        int dr[4] = {0, -1, 0, 1};
        int dc[4] = {-1, 0, 1, 0};
        while(!q.empty())
        {
            vector<int> temp = q.front();
            q.pop();
            int u = temp[0];
            int v = temp[1];
            int d = temp[2];

            ans = max(ans, d);

            for(int i = 0 ; i<4 ; i++)
            {
                int newu = u + dr[i];
                int newv = v + dc[i];

                if(newu<0 || newv<0 || newu>=n || newv>=n || grid[newu][newv]==1)
                {
                    continue;
                }
                grid[newu][newv] = 1;
                int newd = d+1;
                q.push({newu, newv, newd});
            }
        }
        return ans;
    }
};
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