docs(docs): add Approach and Explanation sections with mermaid diagrams to problem pages

apps/docs/content/(linked-list)/206-reverse-linked-list.mdx

@@ -29,6 +29,19 @@ A linked list can be reversed either iteratively or recursively. Could you imple
 - The number of nodes in the list is the range [0, 5000].
 - `-5000 <= Node.val <= 5000`
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["reverseList(head)"]) --> I["prev = None, curr = head"]
+  I --> W{"curr?"}
+  W -- no --> E(["return prev — the old tail is the new head"])
+  W -- yes --> A["next_ = curr.next — save it before overwriting"]
+  A --> B["curr.next = prev — flip the link"]
+  B --> C["prev = curr; curr = next_ — step forward"]
+  C --> W
+```
+
 ## Solution
 
 ```py
@@ -48,3 +61,7 @@ class Solution:
             curr = next_
         return prev
 ```
+
+## Explanation
+
+<YouTube id="G0_I-ZF0S38" title="Reverse Linked List - Iterative AND Recursive - Leetcode 206 - Python" />

apps/docs/content/(linked-list)/21-merge-two-sorted-lists.mdx

@@ -26,6 +26,23 @@ sidebar:
 - `-100 <= Node.val <= 100`
 - Both list1 and list2 are sorted in non-decreasing order.
 
+## Approach
+
+```mermaid
+flowchart TD
+  S(["mergeTwoLists(list1, list2)"]) --> G1{"list1 empty?"}
+  G1 -- yes --> X1(["return list2"])
+  G1 -- no --> G2{"list2 empty?"}
+  G2 -- yes --> X2(["return list1"])
+  G2 -- no --> H["head = the smaller first node; advance that list"]
+  H --> C["current = head"]
+  C --> W{"list1 and list2 both non-empty?"}
+  W -- yes --> P["current.next = the smaller node; advance that list; current = current.next"]
+  P --> W
+  W -- no --> T["current.next = list1 or list2 — append the leftover tail"]
+  T --> E(["return head"])
+```
+
 ## Solution
 
 ```py
@@ -64,3 +81,7 @@ class Solution:
         current.next = list1 or list2
         return head
 ```
+
+## Explanation
+
+<YouTube id="XIdigk956u0" title="Merge Two Sorted Lists - Leetcode 21 - Python" />