<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 isallsame(vector<vector<int>>&grid)
    {
        int compare=grid[0][0];
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if(grid[i][j]!=compare)
                {
                    return false;
                }
            }
        }
        return true;
    }
    int maxDistance(vector<vector<int>>&grid) 
    {
        if(isallsame(grid))
        {
            return -1;
        }
        vector<bool>a(grid[0].size(),false);
        vector<int>b(grid[0].size(),0);
        vector<vector<bool>>pushedinqueue(grid.size(),a);
        vector<vector<int>>distance(grid.size(),b);
        //BFS
        queue<pair<int,int>>visit;
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if(grid[i][j]==1)
                {
                    visit.push({i,j});
                    pushedinqueue[i][j]=true;
                }
            }
        }
        int dist=0;
        while(!visit.empty())
        {
            int n=visit.size();
            for(int i=0;i<n;i++)
            {
                pair<int,int>x=visit.front();
                visit.pop();
                distance[x.first][x.second]=dist;
                if(x.first>0 && !pushedinqueue[x.first-1][x.second])
                {
                    pushedinqueue[x.first-1][x.second]=true;
                    visit.push({x.first-1,x.second});
                }
                if(x.first<(grid.size()-1) && !pushedinqueue[x.first+1][x.second])
                {
                    pushedinqueue[x.first+1][x.second]=true;
                    visit.push({x.first+1,x.second});
                }
                if(x.second>0 && !pushedinqueue[x.first][x.second-1])
                {
                    pushedinqueue[x.first][x.second-1]=true;
                    visit.push({x.first,x.second-1});
                }
                if(x.second<(grid[0].size()-1) && !pushedinqueue[x.first][x.second+1])
                {
                    pushedinqueue[x.first][x.second+1]=true;
                    visit.push({x.first,x.second+1});
                }
            }
            dist++;
        }
        int ans=INT_MIN;
        for(int i=0;i<grid.size();i++)
        {
            for(int j=0;j<grid[0].size();j++)
            {
                if(grid[i][j]==0)
                {
                    ans=max(ans,distance[i][j]);
                }
            }
        }
        return ans;
    }
};

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