<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 {
    vector<int> dr = {0, 0, 1, -1};
    vector<int> dc = {1, -1, 0, 0};

public:
    bool isPos(int r, int c, int n) {
        return (r >= 0 && c >= 0 && r < n && c < n);
    }
    int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        vector<vector<int>> dist(n, vector<int>(n, - 1));
        queue<vector<int>> pq;
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (1 == grid[i][j]) {
                    dist[i][j] = 0;
                    pq.push({i, j});
                }
            }
        }
        int ans = 0;
        while (!pq.empty()) {
            vector<int> vect = pq.front();
            pq.pop();
            int r = vect[0];
            int c = vect[1];
            if (0 == grid[r][c]) {
                ans = max(ans, dist[r][c]);
            }
            for (int i = 0; i < 4; i++) {
                int nr = r + dr[i];
                int nc = c + dc[i];
                if (isPos(nr, nc, n) && -1 == dist[nr][nc]) {
                    dist[nr][nc] = dist[r][c] + 1;
                    pq.push({nr, nc});
                }
            }
        }
        return (ans == 0 ? -1 : ans);
    }
};
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