<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) {
        vector<int> dx = {-1,1,0,0};
        vector<int> dy = {0,0,-1,1};
        int n = grid.size();
        vector<vector<int>> visited(n,vector<int>(n,false));
        vector<vector<int>> distance(n,vector<int>(n,INT_MAX));
        
        queue<vector<int>> q1;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                if(grid[i][j]==1){
                    q1.push({i,j});
                    distance[i][j]=0;
                    visited[i][j]=true;
                }
            }
        }
        while(!q1.empty()){
            vector<int> p1 = q1.front();
            q1.pop();
            visited[p1[0]][p1[1]]=true;
            for(int h=0;h<4;h++){
                int nx = p1[0]+dx[h];
                int ny = p1[1]+dy[h];
                if(nx>=0 && nx<n && ny>=0 && ny<n && !visited[nx][ny] && grid[nx][ny]==0)
                {
                    if(distance[nx][ny]>1+distance[p1[0]][p1[1]]){
                        distance[nx][ny]=1+distance[p1[0]][p1[1]];
                        q1.push({nx,ny});
                    }
                    
                
                    
                }

            }
        }
        int maxm= -1;
        for(int i=0;i<n;i++){
            for(int j=0;j<n;j++){
                maxm = max(maxm,distance[i][j]);
            }
        }
        if(maxm==INT_MAX || maxm==0){
            return -1;
        }
        return maxm;




        
    }
};
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