<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 bfs(vector<vector<int>>& grid, queue<vector<int>>& q){

        int maxD=-1;
        vector<vector<int>> visited(grid.size(), vector<int>(grid[0].size(), false));
        while(!q.empty()){
            int x=q.front()[0];
            int y=q.front()[1];
            int dist=q.front()[2];
            maxD=max(maxD, dist);
            q.pop();

            vector<int> dirx={0,1,-1,0};
            vector<int> diry={1,0,0,-1};

            for(int i=0; i<4; i++){
                int newx=x+dirx[i];
                int newy=y+diry[i];

                if(newx<0 || newy<0 || newx>=grid.size() || newy>=grid[0].size()) continue;
                if(!visited[newx][newy] && grid[newx][newy]==0){
                    visited[newx][newy]=true;
                    q.push({newx,newy,dist+1});
                }

            }
        }
        return maxD;

    }

    int maxDistance(vector<vector<int>>& grid) {

        int maxD=-1;
        int n=grid.size();
        int m=grid[0].size();
        int count0=0;
        queue<vector<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,0});
                    count0++;
                }
            }
        }
        maxD=bfs(grid, q);
        if(count0==n*m) return -1;
        return maxD;

        
    }
};
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