<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>

#include <assert.h>

class Solution {
public:
    int maxDistance(vector<vector<int>>& grid) {
        const int n = grid.size();
        vector< pair<int, int> > cur;
        cur.reserve(n*n);

        for(int i = 0; i < n; ++i) {
            for(int j = 0; j < n; ++j) {
                if(grid[i][j]) 
                    cur.emplace_back(i, j);
            }
        }

        if(cur.empty() || cur.size() == n*n)
            return -1;

        vector< pair<int, int> > cur2;
        cur2.reserve(n*n);
        vector< pair<int, int> >* pcur = &cur;
        vector< pair<int, int> >* next = &cur2;
        
        int val = 1;
        while(true) {

            for(auto& p: *pcur) {
                for(auto p2: vector<pair<int, int>>{ {-1, 0}, {1, 0}, {0, -1}, {0, 1} }) {
                    int ix = p.first + p2.first;
                    int iy = p.second + p2.second;
                    if(ix < 0 || ix >= n || iy < 0 || iy >= n) {
                        continue;
                    }
                    if(grid[ix][iy] == 0) {
                        next->emplace_back(ix, iy);
                        grid[ix][iy] = val + 1;
                    }
                }
            }

            if(next->empty()) {
                return val - 1;  // we haven't written `val` at all into the grid, so the previous step was the last successful
            }

            swap(pcur, next);
            ++val;
            next->clear();
        }

        assert(false);
        return -1;  // never
    }
};
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