<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) {
        list<vector<int>> nexts;

        int ans = -1;

        for (int i = 0; i < grid.size(); i++) {
            for (int j = 0; j < grid.size(); j++) {
                if (grid[i][j] == 1) {
                    nexts.push_back({i, j, 0});
                }
            }
        }

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

        vector<pair<int, int>> directions {{-1, 0}, {1,0}, {0,-1}, {0,1}};

        while (!nexts.empty()) {
            int row = nexts.front()[0];
            int col = nexts.front()[1];
            int dist = nexts.front()[2];
            nexts.pop_front();

            ans = dist;
            for (const auto& dir : directions) {
                int new_row = row + dir.first;
                int new_col = col + dir.second;

                if (new_row >= 0 && new_row < grid.size() && new_col >= 0 && new_col < grid[0].size()) {
                    if (grid[new_row][new_col] == 0) {
                        grid[new_row][new_col] = 1;
                        nexts.push_back({new_row, new_col, dist+1});
                    }
                }
            }
        }

        return ans;
    }
};
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