<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:
    bool check(int i,int j,int n) {
        if(i<0 or j<0 or i==n or j==n) return false;
        return true;
    }

    int maxDistance(vector<vector<int>>& grid) {
        int res = 0;
        queue<vector<int>> q;
        int n = grid.size();
        vector<vector<int>> dist(n,vector<int> (n,0));
        vector<vector<bool>> visited(n,vector<bool> (n,false));
        vector<vector<int>> moves = {{1,0},{-1,0},{0,1},{0,-1}};
        for(int i = 0 ; i<n ; i++) {
            for(int j = 0 ; j<n ; j++) {
                if(grid[i][j]==1) {
                    visited[i][j] = true;
                    q.push({i,j,0});
                }
            }
        }
        while(!q.empty()) {
            int i = q.front()[0];
            int j = q.front()[1];
            int currDist = q.front()[2];
            if(currDist>res) res = currDist;
            q.pop();
            for(auto &x:moves) {
                int in = i + x[0], jn = j + x[1];
                if(check(in,jn,n) and !visited[in][jn]) {
                    visited[in][jn] = true;
                    dist[in][jn] = currDist + 1;
                    q.push({in,jn,currDist+1});
                }
            }
        }
        return res==0?-1:res;
    }
};
42168