<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();
       queue<pair<int,int>> q;
       for(int i=0;i<n;i++){
            for(int j=0;j<m;j++){
                if(grid[i][j] == 1){
                    q.push({i,j});
                    grid[i][j] = -1;
                } 
            }
       }
       int d = 0;
       while(!q.empty()){
         d++;
         int sz = q.size();
         for(int i=0;i<sz;i++){
            auto p = q.front();
            int x = p.first;
            int y = p.second;
            q.pop();
            vector<int> delr = {-1,0,1,0};
            vector<int> delc = {0,1,0,-1};
            for(int k=0;k<4;k++){
                int nx = x + delr[k];
                int ny = y + delc[k];
                if(nx < 0 || nx >=n || ny < 0 || ny >= m || grid[nx][ny] != 0){
                    continue;
                }
                grid[nx][ny] = d;
                q.push({nx,ny});
            }
         }
       }
       int ans = -1e9;
       for(int i=0;i<n;i++){
        for(int j = 0;j<m;j++){
            if(grid[i][j] != -1){
                ans = max(ans, grid[i][j]);
            }
        }
       }
       if(ans == 0 || ans == -1e9){
        return -1;
       } 
       return ans;
    }
};
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