<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:

    bool isvalid(int x, int y, vector<vector<int>>& grid, set<pair<int,int>> &visited){
        int n = grid.size(); int m = grid[0].size();
        if(x>=0 && x<n && y>=0 && y<m && grid[x][y]==0 && visited.find({x,y})==visited.end()){
            return true;
        }
        return false;
    }

    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});
            }
        }

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

        set<pair<int,int>> visited;
        int drow[] = {1,-1,0,0};
        int dcol[] = {0,0,1,-1};

        int dist = 0;
        
        while(!q.empty()){

            int s = q.size();
            for(int i=0; i<s; i++){
                int curx = q.front().first;
                int cury = q.front().second;
                q.pop();
                for(int k=0; k<4; k++){
                    int nx = curx + drow[k];
                    int ny = cury + dcol[k];

                    if(isvalid(nx,ny,grid,visited)){ // valid and is 0 and isnt touched before
                        visited.insert({nx,ny});
                        q.push({nx,ny});
                    }

                }

            }
            dist++;
        }

        return dist-1;
    }
};
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