docs: rename category folders to remove parentheses and update heap meta title

apps/docs/content/graph-dfs/200-number-of-islands.mdx

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+---
+title: '200. Number of Islands'
+description: Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands
+sidebar:
+  label: 'Number of Islands'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+### Example 1:
+- Input: `grid = [ ["1","1","1","1","0"], ["1","1","0","1","0"], ["1","1","0","0","0"], ["0","0","0","0","0"] ]`
+- Output: `1`
+
+### Example 2:
+- Input: `grid = [ ["1","1","0","0","0"], ["1","1","0","0","0"], ["0","0","1","0","0"], ["0","0","0","1","1"] ]`
+- Output: `3`
+
+### Constraints:
+
+- `m == grid.length`
+- `n == grid[i].length`
+- `1 <= m, n <= 300`
+- grid[i][j] is '0' or '1'.
+
+## Solution
+
+```py
+class Solution:
+    def numIslands(self, grid: List[List[str]]) -> int:
+        rows, cols = len(grid), len(grid[0])
+
+        def dfs(r, c):
+            if r < 0 or r >= rows or c < 0 or c >= cols or grid[r][c] != "1":
+                return
+            else:
+                grid[r][c] = "0"
+                dfs(r, c + 1)  # right
+                dfs(r + 1, c)  # bottom
+                dfs(r, c - 1)  # left
+                dfs(r - 1, c)  # top
+
+        num_islands = 0
+        for r in range(rows):
+            for c in range(cols):
+                if grid[r][c] == "1":
+                    num_islands += 1
+                    dfs(r, c)
+
+        return num_islands
+```