Sync LeetCode submission Runtime - 242 ms (60.36%), Memory - 20.1 MB (44.37%)

Author: jimit105

0200-number-of-islands/README.md

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+<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>

0200-number-of-islands/solution.py

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+# Approach 1: DFS
+
+# m = no. of rows, n = no. of columns
+# Time: O(m * n)
+# Space: O(m * n)
+
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        if not grid:
+            return 0
+
+        num_islands = 0
+        for r in range(len(grid)):
+            for c in range(len(grid[0])):
+                if grid[r][c] == "1":
+                    self.dfs(grid, r, c)
+                    num_islands += 1
+
+        return num_islands
+
+    def dfs(self, grid, r, c):
+        if r < 0 or c < 0 or r >= len(grid) or c >= len(grid[0]) or grid[r][c] != "1":
+            return
+
+        grid[r][c] = "0"
+
+        self.dfs(grid, r - 1, c)
+        self.dfs(grid, r + 1, c)
+        self.dfs(grid, r, c - 1)
+        self.dfs(grid, r, c + 1)
+
+