<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 res = 0;
        queue<pair<int,int>> q;
        
        //find all lands and add them to q (as start points of the bfs) (first step)
        for (int i = 0; i < grid.size(); i++)
            for (int j = 0; j < grid.size(); j++)
                if (grid[i][j] == 1)
                    q.push({i,j});
    
               while(!q.empty()){
            pair<int,int> curr= q.front();
            q.pop();
            
            //4 directions
            vector<pair<int,int>> dirs {{0,1},{0,-1},{1,0},{-1,0}};
            
            //check for 4 directions
            for (auto dir : dirs){
                int i = curr.first + dir.first;
                int j = curr.second + dir.second;
                
                //avoid getting i and j outside the grid && check if grid[i][j] == 0
                if (i >= 0 && i < grid.size() && j >= 0 && j < grid.size() && grid[i][j] == 0){
                    grid[i][j] = grid[curr.first][curr.second] + 1;
                    q.push({i,j});
                    res = max(res,grid[i][j]);
                }
            }
        }       
        return res - 1;        
    }
};
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