<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) {
        queue<pair<int, pair<int, int>>> fifo;
        vector<vector<bool>> visited(grid.size(), vector<bool>(grid[0].size(), false));
        
        for(int i = 0; i < grid.size(); i++){
            for(int j = 0; j < grid[i].size(); j++){
                if(grid[i][j] == 1) {
                    fifo.push({0, {i, j}});
                    visited[i][j] = true;
                }
            }
        }

        if(fifo.size() == 0 || fifo.size() == grid.size() * grid[0].size()) return -1;
        int maxDistance = 0;
        while(!fifo.empty()){
            int dist = fifo.front().first;
            int x = fifo.front().second.first;
            int y = fifo.front().second.second;
            fifo.pop();

            maxDistance = max(maxDistance, dist);

            vector<pair<int, int>> directions = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
            for(auto dir : directions){
                int newX = x + dir.first;
                int newY = y + dir.second;

                if(newX >= 0 && newY >= 0 && newX < grid.size() && newY < grid[0].size() && grid[newX][newY] == 0 && !visited[newX][newY]){
                    fifo.push({dist + 1, {newX, newY}});
                    visited[newX][newY] = true;
                }
            }
        }
        return maxDistance;
    }
};

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