<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 rows = grid.size(), cols = grid[0].size();
        deque<tuple<int,int>> q;
        set<tuple<int,int>> seen;

        for (int r=0; r<rows; ++r)
            for (int c=0; c<cols; ++c) {
                if (grid[r][c]) {
                    seen.insert({r,c});
                    q.push_back({r,c});
                }
            }
        if (q.empty() || q.size() == rows*cols) return -1;

        int step = -1;
        while (!q.empty()) {
            int size = q.size();
            for (int i=0; i<size; ++i) {
                auto[r,c] = q.front();
                q.pop_front();
                
                vector<tuple<int,int>> move = {{r+1,c},{r-1,c},{r,c+1},{r,c-1}};
                for (const auto& [nr,nc]: move) {
                    if (0<=nr && nr<rows && 0<=nc && nc<cols && seen.find({nr,nc})==seen.end()) {
                        seen.insert({nr,nc});
                        q.push_back({nr,nc});
                    }
                }
            }
            step += 1;
        }
        return step;
    }
};
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