<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) {
        std::ios_base::sync_with_stdio(false);
 std::cin.tie(NULL);
        int n=grid.size();
        int m=grid[0].size();
        vector<vector<int>>dist(n,vector<int>(m,1e9));
        queue<pair<int,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({0,{i,j}});
                    dist[i][j] = 0;
                }
            }
        }
        int dr[]={-1,0,1,0};
        int dc[]={0,1,0,-1};
        int maxd=-1;
        while(!q.empty()){
            int dis=q.front().first;
            int r=q.front().second.first;
            int c=q.front().second.second;
            q.pop();
            for(int i=0;i<4;i++){
                int nr=r+dr[i];
                int nc=c+dc[i];
                if(nr>=0 && nr<n && nc>=0 && nc<m && grid[nr][nc]==0 && dist[nr][nc]>1+dis){
                        dist[nr][nc]=1+dis;
                        q.push({1+dis,{nr,nc}});
                        maxd=max(maxd,1+dis);   
                }
            }

        }
       if(maxd==-1) return -1;
       return maxd; 
    }
};
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