<p>Given an <code>m x n</code> 2D binary grid <code>grid</code> which represents a map of <code>&#39;1&#39;</code>s (land) and <code>&#39;0&#39;</code>s (water), return <em>the number of islands</em>.</p>

<p>An <strong>island</strong> is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;]
]
<strong>Output:</strong> 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> grid = [
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;1&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;0&quot;,&quot;0&quot;],
  [&quot;0&quot;,&quot;0&quot;,&quot;0&quot;,&quot;1&quot;,&quot;1&quot;]
]
<strong>Output:</strong> 3
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == grid.length</code></li>
	<li><code>n == grid[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 300</code></li>
	<li><code>grid[i][j]</code> is <code>&#39;0&#39;</code> or <code>&#39;1&#39;</code>.</li>
</ul>


class Solution {
public:
int numIslands(vector<vector<char>>& grid) {
    int m=grid.size();
    int n=grid[0].size();
    function<int(int)> Find;
    unordered_map<int,int> Parents;
    unordered_map<int,int> Rank;
    Find=[&grid,&Find,&Parents](int i)->int{
        return Parents[i]==i?i:Find(Parents[i]);
    };
    int cnt=0;
    auto Initialize=[&](){
        for(int i=0;i<grid.size();i++)
            for(int j=0;j<grid[0].size();j++)
            {
                if(grid[i][j]=='1')
                {
                    Parents[i*n+j]=i*n+j;
                    cnt++;
                }
                Rank[i*n+j]=1;
            }
    };
    Initialize();
    auto Union=[&](int x, int y){
        int rootX=Find(x);
        int rootY=Find(y);
        if(rootX!=rootY){
        if(Rank[rootX]>Rank[rootY])
            Parents[rootY]=rootX;
        else if(Rank[rootX]<Rank[rootY])
            Parents[rootX]=rootY;
        else
        {
            Parents[rootY]=rootX;
            Rank[rootX]++;
        }
        cnt--;
        }
    };
    for(int i=0;i<grid.size();i++)
        for(int j=0;j<grid[0].size();j++)
        {
            if(grid[i][j]=='1')
            {
                if(i+1<m&&grid[i+1][j]=='1')
                {
                    Union(n*i+j,n*(i+1)+j);
                }
                if(j+1<n&&grid[i][j+1]=='1')
                {
                    Union(n*i+j,n*i+j+1);
                }
            }
        }
    return cnt;
}
};
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