<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 n = grid.size();
        int water = 0;
        queue<pair<int, int>> q;
        for(int i=0; i<n; i++){
            for(int j=0; j<n; j++){
                if(grid[i][j]) q.push({i, j});
                else water++;
            }
        }
        int dx[] = {-1,1,0,0};
        int dy[] = {0,0,-1,1};
        int distance = 0;
        int maxd = 0;
        if(q.size() == 0 || water == 0) return -1;
        while(!q.empty()){
            int size = q.size();
            distance++;
            while(size--){
                int currRow = q.front().first;
                int currCol = q.front().second;
                q.pop();

                for(int i=0; i<4; i++){
                    int nextRow = currRow + dx[i];
                    int nextCol = currCol + dy[i];
                    if(nextRow >= 0 && nextCol >= 0 && nextRow < n && nextCol < n && grid[nextRow][nextCol]==0){
                        q.push({nextRow, nextCol});
                        grid[nextRow][nextCol] = 1;
                        maxd = max(maxd, distance);
                    }
                }
            }
        }
        return maxd;
    }
};
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