<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 R = grid.size(), C = grid[0].size();
        queue<pair<int, int>> q;
        vector<vector<bool>> seen(R, vector<bool>(C, false));
        vector<vector<int>> distances(R, vector<int>(C, INT_MAX));
        int ans = 0;
        int dirs[5] = {-1,0,1,0,-1};

        for (int r=0; r<R; r++){
            for (int c=0; c<C; c++){
                if (grid[r][c]) {q.push(make_pair(r,c));
                distances[r][c] = 0;}
            }
        }
        if (q.empty() || q.size() == R*C) return -1;
        while (!q.empty()){
            int x = q.front().first;
            int y = q.front().second;
            q.pop();
            if (seen[x][y]) continue; 
            seen[x][y] = true;
            for (int i=0; i<4; i++){
                int nx = x + dirs[i], ny = y + dirs[i+1];
                if (nx >=0 && nx<R && ny>=0 && ny<C && !seen[nx][ny]){
                    distances[nx][ny] = min(distances[x][y] + 1, distances[nx][ny]);
                    ans = max(ans, distances[nx][ny]);
                    q.push(make_pair(nx, ny));
                }
            }
        }
        return ans;
    }
};
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