<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) {
        queue<pair<int,pair<int,int>>> pq;
        vector<vector<int>> dist(grid.size()+1, vector<int> (grid[0].size()+1, INT_MAX));
        vector<vector<bool>> visited(grid.size()+1, vector<bool> (grid[0].size()+1, false));
        int dRow[] = {-1,0,1,0};
        int dCol[] = {0,1,0,-1};
        bool no_water= true;

        for(int i =0; i<grid.size(); i++){
            for(int j =0; j<grid[0].size(); j++){
                if(grid[i][j]==1)
                {
                    visited[i][j] = true;
                    pq.push({0,{i,j}});
                    dist[i][j] =0;
                }
                if(grid[i][j]==0)
                    no_water = false;
            }
        }
        if(pq.empty() || no_water )
            return -1;
        while(!pq.empty()){
            pair<int, pair<int,int>> p = pq.front();
            int d = p.first;
            int x = p.second.first;
            int y = p.second.second;
            pq.pop();
            for(int i =0; i<4; i++){
                int nx= x+dRow[i];
                int ny= y+dCol[i];
                if(nx>=0 && nx<grid.size() && ny>=0 && ny <grid[0].size() && grid[nx][ny]==0)
                {
                    if(dist[nx][ny] > d+1)
                    {
                        dist[nx][ny] = d+1;
                        pq.push({dist[nx][ny],{nx,ny}});
                    }
                }
            }
        }

        int res = INT_MIN;
        for(int i =0; i< grid.size(); i++){
            for(int j = 0; j< grid[0].size(); j++){
                res = max(res, dist[i][j]);
            }
        }
        return res;
    }
};
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