<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 max = 0;
        std::queue<std::pair<int, int>> my_queue;
        for (int r = 0; r < grid.size(); ++r) {
          for (int c = 0; c < grid[0].size(); ++c) {
            if (grid[r][c] == 1) {
              my_queue.push(std::make_pair(r, c));
            }
          }
        }

        while (!my_queue.empty()) {
          auto cur = my_queue.front();
          my_queue.pop();
          int r = cur.first;
          int c = cur.second;
          int dist = grid[r][c];
          if (dist > max) max = dist;

          // up
          visit(grid, my_queue, r-1, c, dist);
          // bottom
          visit(grid, my_queue, r+1, c, dist);
          // left
          visit(grid, my_queue, r, c-1, dist);
          // right
          visit(grid, my_queue, r, c+1, dist);
        }

        return max == 1 ? -1 : max-1;
    }

    void visit(vector<vector<int>>& grid, std::queue<std::pair<int, int>>& my_queue, int r, int c, int dist) {
      if (r < 0 || c < 0 || r >= grid.size() || c >= grid[0].size() || grid[r][c] != 0) {
        return;
      }
      grid[r][c] = dist + 1;
      my_queue.push(std::make_pair(r,c));
    }
};
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