<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:
    vector<int>g[105];
    int dist[1001][1001];
    int vis[1001][1001];
    int n,m;
    bool f=true,ff=true;;

bool isValid(int x, int y) {
    if(x<0 || x>=n || y<0 || y>=m) return false;
    if(vis[x][y]==1) return false;
    return true;
}

int dx[4]={-1,0,1,0};
int dy[4]={0,1,0,-1};

void bfs ()
{
    queue<pair<int,int>>q;
   

    for(int i=0; i<n; i++) {
        for(int j=0; j<m; j++) {
            if(g[i][j]==1) {
                q.push({i,j});
                dist[i][j]=0;
                f=false;
            } 
            else ff=false;
        }
    }

    while(!q.empty()) {
        int cx=q.front().first;
        int cy=q.front().second;
        q.pop();
        for(int i=0; i<4; i++) {
            if(isValid(cx+dx[i],cy+dy[i])) {
                int nx=cx+dx[i];
                int ny=cy+dy[i];
                vis[nx][ny]=1;
                q.push({nx,ny});
                if(dist[nx][ny]>dist[cx][cy]+1)
                dist[nx][ny]=dist[cx][cy]+1;
            }
        }
    }

}

    int maxDistance(vector<vector<int>>& grid) {
        n=grid.size();
        m=grid[0].size();
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) g[i].push_back(j);
        }

        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) g[i][j]=grid[i][j];
        }
        memset(dist,1,sizeof(dist));
        bfs();
        int mx=0;
        for(int i=0; i<n; i++) {
            for(int j=0; j<m; j++) mx=max(mx,dist[i][j]);
        }

        if(f || ff) return -1;
        else return mx;

    }
};
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