<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:
    vector<vector<int>> dir = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}};
    int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        int ans = -1;
        queue<pair<pair<int, int>, int>> q;
        set<pair<int,int>>vis;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1) {
                    q.push({{i, j}, 0});
                    vis.insert({i, j});
                }
            }
        }
        if (q.size() == 0 or q.size() == n * n)//only land or only water
            return -1;

        // bfs
        while (q.size()) {
            auto p = q.front();
            q.pop();
            int x = p.first.first;
            int y = p.first.second;
            ans = max(ans, p.second);
            for (int i = 0; i < 4; i++) {
                int nx = x + dir[i][0];
                int ny = y + dir[i][1];
                if (nx >= 0 and ny >= 0 and nx < n and ny < n and
                    vis.count({nx, ny}) == 0) {
                    q.push({{nx, ny}, p.second + 1});
                    vis.insert({nx, ny});
                }
            }
        }
        return ans;
    }
};
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