<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();
        int cols = grid[0].size();
        vector<vector<bool>> visited(rows, vector<bool>(cols, false));
        queue<pair<int, int>> q;

        for (int i = 0; i < rows; ++i) {
            for (int j = 0; j < cols; ++j) {
                if (grid[i][j] == 1) {
                    q.push({i, j});
                    visited[i][j] = true;
                }
            }
        }

        if (q.empty() || q.size() == rows * cols)
            return -1;

        int maxDist = -1;

        while (!q.empty()) {
            int size = q.size();
            maxDist++;
            while (size--) {
                pair<int, int> newPair = q.front();
                q.pop();
                int i = newPair.first;
                int j = newPair.second;

                if (i - 1 >= 0 && !visited[i - 1][j] && grid[i - 1][j] == 0) {
                q.push({i - 1, j});
                visited[i - 1][j] = true;
            }
            if (j - 1 >= 0 && !visited[i][j - 1] && grid[i][j - 1] == 0) {
                q.push({i, j - 1});
                visited[i][j - 1] = true;
            }
            if (j + 1 < cols && !visited[i][j + 1] && grid[i][j + 1] == 0) {
                q.push({i, j + 1});
                visited[i][j + 1] = true;
            }
            if (i + 1 < rows && !visited[i + 1][j] && grid[i + 1][j] == 0) {
                q.push({i + 1, j});
                visited[i + 1][j] = true;
            }
        }
        }
        return maxDist;
    }
};
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