<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 dx[4] = {1,-1,0,0};
    int dy[4] = {0,0,1,-1};
    int maxDistance(vector<vector<int>>& grid) {
        int n = grid.size();
        int vis[n+1][n+1];
        memset(vis , -1, sizeof vis);

        queue<array<int,3>>q;
        int maxDis = -1;

        for(int i = 0 ; i<n ; i++){
            for(int j = 0; j < n ; j++){
                if(grid[i][j]) q.push({i,j,0});
            }
        }

        while(!q.empty()){
                    
            int x = q.front()[0];
            int y = q.front()[1];
            int dis = q.front()[2];

            q.pop();

            if(x>=n || y>=n || x<0 || y<0) continue;

            if(vis[x][y]!=-1) continue;
            vis[x][y]=1;
            
            if(grid[x][y]==0)
            maxDis=max(maxDis,dis);

            for(int i = 0 ; i < 4 ; i++){
                q.push({x+dx[i],y+dy[i],dis+1});
            }

        
        }
        return maxDis;
        
    
    }
};
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