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

apps/docs/content/graph-dfs/133-clone-graph.mdx

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+---
+title: '133. Clone Graph'
+description: Given a reference of a node in a connected undirected graph
+sidebar:
+  label: 'Clone Graph'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Hash Table</Badge>
+
+### Example 1:
+- Input: `adjList = [[2,4],[1,3],[2,4],[1,3]]`
+- Output: `[[2,4],[1,3],[2,4],[1,3]]`
+- Explanation: There are `4` nodes in the graph. 1st `node` (val = 1)'s neighbors are 2nd `node` (val = `2`) and 4th `node` (val = `4`). 2nd `node` (val = 2)'s neighbors are 1st `node` (val = `1`) and 3rd `node` (val = `3`). 3rd `node` (val = 3)'s neighbors are 2nd `node` (val = `2`) and 4th `node` (val = `4`). 4th `node` (val = 4)'s neighbors are 1st `node` (val = `1`) and 3rd `node` (val = `3`).
+
+### Example 2:
+- Input: `adjList = [[]]`
+- Output: `[[]]`
+- Explanation: Note that the input contains one empty list. The graph consists of only one `node` with val = `1` and it does not have any neighbors.
+
+### Example 3:
+- Input: `adjList = []`
+- Output: `[]`
+- Explanation: This an empty graph, it does not have any nodes.
+
+### Constraints:
+
+- The number of nodes in the graph is in the range [0, 100].
+- `1 <= Node.val <= 100`
+- Node.val is unique for each node.
+- There are no repeated edges and no self-loops in the graph.
+- The Graph is connected and all nodes can be visited starting from the given node.
+
+## Solution
+
+```py
+"""
+# Definition for a Node.
+class Node:
+    def __init__(self, val = 0, neighbors = None):
+        self.val = val
+        self.neighbors = neighbors if neighbors is not None else []
+"""
+
+from typing import Optional
+
+
+class Solution:
+    def cloneGraph(self, node: Optional["Node"]) -> Optional["Node"]:
+        oldToNew = {}
+
+        def dfs(node):
+            if node in oldToNew:
+                return oldToNew[node]
+
+            copy = Node(node.val)
+            oldToNew[node] = copy
+
+            for n in node.neighbors:
+                copy.neighbors.append(dfs(n))
+            return copy
+
+        return dfs(node) if node else None
+```

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
+```

apps/docs/content/graph-dfs/417-pacific-atlantic-water-flow.mdx

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+---
+title: '417. Pacific Atlantic Water Flow'
+description: There is an m x n rectangular island that borders both the Pacific Ocean and Atlantic Ocean. The Pacific Ocean touches the island's left and top edges, and the Atlantic Ocean touches the island's right and bottom edges
+sidebar:
+  label: 'Pacific Atlantic Water Flow'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Array</Badge>
+
+### Example 1:
+- Input: `heights = [[1,2,2,3,5],[3,2,3,4,4],[2,4,5,3,1],[6,7,1,4,5],[5,1,1,2,4]]`
+- Output: `[[0,4],[1,3],[1,4],[2,2],[3,0],[3,1],[4,0]]`
+- Explanation: The following cells can flow to the Pacific and Atlantic oceans, as shown below: `[0,4]`: `[0,4]` -> Pacific Ocean `[0,4]` -> Atlantic Ocean `[1,3]`: `[1,3]` -> `[0,3]` -> Pacific Ocean `[1,3]` -> `[1,4]` -> Atlantic Ocean `[1,4]`: `[1,4]` -> `[1,3]` -> `[0,3]` -> Pacific Ocean `[1,4]` -> Atlantic Ocean `[2,2]`: `[2,2]` -> `[1,2]` -> `[0,2]` -> Pacific Ocean `[2,2]` -> `[2,3]` -> `[2,4]` -> Atlantic Ocean `[3,0]`: `[3,0]` -> Pacific Ocean `[3,0]` -> `[4,0]` -> Atlantic Ocean `[3,1]`: `[3,1]` -> `[3,0]` -> Pacific Ocean `[3,1]` -> `[4,1]` -> Atlantic Ocean `[4,0]`: `[4,0]` -> Pacific Ocean `[4,0]` -> Atlantic Ocean Note that there are other possible paths for these cells to flow to the Pacific and Atlantic oceans.
+
+### Example 2:
+- Input: `heights = [[1]]`
+- Output: `[[0,0]]`
+- Explanation: The water can flow from the only cell to the Pacific and Atlantic oceans.
+
+### Constraints:
+
+- `m == heights.length`
+- `n == heights[r].length`
+- `1 <= m, n <= 200`
+- `0 <= heights[r][c] <= 10^5`
+
+## Solution
+
+```py
+class Solution:
+    def pacificAtlantic(self, heights: list[list[int]]) -> list[list[int]]:
+```

apps/docs/content/graph-dfs/meta.ts

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+import { defineMeta } from "blume";
+
+export default defineMeta({ title: "Graph DFS", order: 16 });