<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 m=grid[0].size();
        int ans=0,count=0;
        for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                count+=grid[i][j];
            }
        }
        if(count==0||count==n*m) return -1;
        while(1){
            vector<vector<int>> qt;
            qt=grid;
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    if(grid[i][j]==0){
                        if(i>0) if(grid[i-1][j]==1) qt[i][j]=1;
                        if(j>0) if(grid[i][j-1]==1) qt[i][j]=1;
                        if(i<n-1) if(grid[i+1][j]==1) qt[i][j]=1;
                        if(j<m-1) if(grid[i][j+1]==1) qt[i][j]=1;
                    }
                }
            }
            ans++;
            grid=qt;
            int cnt=0;
            for(int i=0;i<n;i++){
                for(int j=0;j<m;j++){
                    cnt+=grid[i][j];
                }
            }
            if(cnt==0||cnt==n*m) break;
        }
        return ans;
    }
};
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