<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 dx[]={-1, 0, 1, 0};
        int dy[]={0, -1, 0, 1};
        vector<vector<bool>> visited(n, vector<bool>(n, false));
        queue<vector<int>> q;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                if (grid[i][j] == 1){
                    q.push({i, j, 0});
                }
            }
        }
        if(q.size()==0) return -1;

        int dist=INT_MIN;

        while(!q.empty()){
            auto frontNode=q.front();
            q.pop();
            int x=frontNode[0];
            int y=frontNode[1];
            int curDist=frontNode[2];
            dist=max(curDist, dist);
            for(int i=0; i<4; i++){
                int newX=dx[i]+x, newY=dy[i]+y;
                if(newX>=0 && newY>=0 && newX<n && newY<n && !visited[newX][newY] && grid[newX][newY]==0){
                    // cout<<"...."<<x<<" "<<y<<" "<<curDist<<endl;
                    int newDist=curDist+abs(newX-x)+ abs(newY-y);
                    // cout<<newX<<" "<<newY<<endl;
                    // cout<<newDist<<endl<<endl;;
                    q.push({newX, newY, newDist});
                    visited[newX][newY]=true;
                }
            }
        }
        return dist==0?-1:dist;
    }
};
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