<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:
    def numIslands(self, grid: List[List[str]]) -> int:
        visited = {}
        row = len(grid)
        col = len(grid[0])
        numberOfIslands =0

        for r in range(len(grid)):
            for c in range(len(grid[r])):
                if not visited.get((r,c)) and grid[r][c]!="0":
                    self._dfs(r,c,row,col,grid,visited)
                    numberOfIslands+=1
        return numberOfIslands
    
    def _dfs(self,r,c,row,col,grid,visited):
        if r<0 or c<0 or r>=row or c>=col:
            return 
        elif grid[r][c] == "0":
            return 
        visited[(r,c)] = True
        if not visited.get((r+1,c)):
            self._dfs(r+1,c,row,col,grid,visited)
        if not visited.get((r,c+1)):
            self._dfs(r,c+1,row,col,grid,visited)
        if not visited.get((r-1,c)):
            self._dfs(r-1,c,row,col,grid,visited)    
        if not visited.get((r,c-1)):
            self._dfs(r,c-1,row,col,grid,visited)
        return 
        


        
    
       



        
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