<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();
        vector<vector<int>> vis(n,vector<int>(n,0));

        queue<pair<pair<int,int>,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});
                    vis[i][j] = 1;
                }
            }
        }

        if(q.size() == 0) return -1;
        if(q.size() == n*n) return -1;

        int max_d = INT_MIN;

        while(!q.empty()){
            pair<pair<int,int>,int> node = q.front();
            q.pop();
            int r = node.first.first;
            int c = node.first.second;
            int d = node.second;
            max_d = max(max_d,d);

            vector<pair<int,int>> dir = {{-1,0},{1,0},{0,-1},{0,1}};
            for(auto it: dir){
                int nr = r + it.first;
                int nc = c + it.second;

                if(nr >= 0 && nc >=0 && nr < n && nc < n 
                && grid[nr][nc] == 0 && vis[nr][nc]!=1 )
                {
                    vis[nr][nc] = 1;
                    q.push({{nr,nc},d+1});
                    
                }
            }
        }
        return max_d;


    }
};
41556