<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) {
        // if grid is empty we should return -1
        if (grid.size() == 0) { return -1; }

        std::vector<std::pair<int, int>> direcs;
        direcs.push_back({-1, 0});
        direcs.push_back({1, 0});
        direcs.push_back({0, 1});
        direcs.push_back({0, -1});


        std::set<std::pair<int, int>> visited;
        std::queue<std::pair<int, int>> q;
        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid[0].size(); j++) {
                if (grid[i][j] == 1) { 
                    q.push({i, j}); 
                    visited.insert({i, j});
                }
            }
        }

        if (q.size() == (grid.size() * grid[0].size())) { return -1; }

        int res = -1;
        while (!q.empty()) {
            int size = q.size();
            for (int i = 0; i < size; i++) {
                auto cur = q.front(); q.pop();
                
                for (auto direc : direcs) {
                    int r = direc.first + cur.first;
                    int c = direc.second + cur.second;

                    if (r > -1 && r < grid.size() &&
                        c > -1 && c < grid[0].size() &&
                        grid[r][c] == 0 &&
                        visited.find({r, c}) == visited.end()) {
                        q.push({r, c});
                        visited.insert({r, c});
                    }
                }
            }
            res++;
        }
        return res;
    
    }
};
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