docs: add new problem documentation pages and update redirects

apps/docs/content/(02-nodal-grid)/linked-list/19_remove-nth-node-from-end-of-list.mdx

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+---
+title: '19. Remove Nth Node From End of List'
+description: Given the head of a linked list, remove the n^th node from the end of the list and return its head
+icon: dot
+topics:
+  - { name: "Linked List", slug: "linked-list" }
+  - { name: "Two Pointers", slug: "two-pointers" }
+issue: "https://github.com/prdlk/leetcode/issues/80"
+sidebar:
+  label: 'Remove Nth Node From End of List'
+  badge: 'Medium'
+---
+
+::::warning
+Could you do this in one pass?
+::::
+
+### Example 1:
+- Input: `head = [1,2,3,4,5], n = 2`
+- Output: `[1,2,3,5]`
+
+### Example 2:
+- Input: `head = [1], n = 1`
+- Output: `[]`
+
+### Example 3:
+- Input: `head = [1,2], n = 1`
+- Output: `[1]`
+
+### Constraints:
+
+- The number of nodes in the list is sz.
+- `1 <= sz <= 30`
+- `0 <= Node.val <= 100`
+- `1 <= n <= sz`
+
+## Solution
+
+```py
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def removeNthFromEnd(self, head: ListNode | None, n: int) -> ListNode | None:
+        dummy = ListNode(0)
+        dummy.next = head
+
+        fast = dummy
+        for _ in range(n):
+            if fast.next is None:
+                return head
+            fast = fast.next
+
+        slow = dummy
+        while fast.next is not None:
+            slow = slow.next
+            fast = fast.next
+
+        slow.next = slow.next.next
+
+        return dummy.next
+```

apps/docs/content/(02-nodal-grid)/linked-list/82_remove-duplicates-from-sorted-list-ii.mdx

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+---
+title: '82. Remove Duplicates from Sorted List II'
+description: You are given the head of a sorted linked list
+icon: dot
+topics:
+  - { name: "Linked List", slug: "linked-list" }
+  - { name: "Two Pointers", slug: "two-pointers" }
+issue: "https://github.com/prdlk/leetcode/issues/84"
+sidebar:
+  label: 'Remove Duplicates from Sorted List II'
+  badge: 'Medium'
+---
+
+### Example 1:
+- Input: `head = [1,2,3,3,4,4,5]`
+- Output: `[1,2,5]`
+
+### Example 2:
+- Input: `head = [1,1,1,2,3]`
+- Output: `[2,3]`
+
+### Constraints:
+
+- The number of nodes in the list is in the range [0, 300].
+- `-100 <= Node.val <= 100`
+- The list is guaranteed to be sorted in ascending order.
+
+## Solution
+
+```py
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def deleteDuplicates(self, head: ListNode | None) -> ListNode | None:
+        sentinel = ListNode(0, head)
+        pred = sentinel
+
+        while head:
+            if head.next and head.val == head.next.val:
+                while head.next and head.val == head.next.val:
+                    head = head.next
+
+                pred.next = head.next
+            else:
+                pred = pred.next
+
+            head = head.next
+
+        return sentinel.next
+```

apps/docs/content/(02-nodal-grid)/linked-list/876_middle-of-the-linked-list.mdx

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+---
+title: '876. Middle of the Linked List'
+description: Given the head of a singly linked list, return the middle node of the linked list
+icon: dot
+topics:
+  - { name: "Linked List", slug: "linked-list" }
+  - { name: "Two Pointers", slug: "two-pointers" }
+issue: "https://github.com/prdlk/leetcode/issues/82"
+sidebar:
+  label: 'Middle of the Linked List'
+  badge: 'Easy'
+---
+
+### Example 1:
+- Input: `head = [1,2,3,4,5]`
+- Output: `[3,4,5]`
+- Explanation: The middle node of the list is node `3`.
+
+### Example 2:
+- Input: `head = [1,2,3,4,5,6]`
+- Output: `[4,5,6]`
+- Explanation: Since the list has two middle nodes with values `3` and `4`, we return the second one.
+
+### Constraints:
+
+- The number of nodes in the list is in the range [1, 100].
+- `1 <= Node.val <= 100`
+
+## Solution
+
+```py
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def middleNode(self, head: ListNode | None) -> ListNode | None:
+        slow = fast = head
+        while fast and fast.next:
+            fast = fast.next.next
+            slow = slow.next
+
+        return slow
+```

apps/docs/content/(02-nodal-grid)/linked-list/92_reverse-linked-list-ii.mdx

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+---
+title: '92. Reverse Linked List II'
+description: Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list
+icon: dot
+topics:
+  - { name: "Linked List", slug: "linked-list" }
+issue: "https://github.com/prdlk/leetcode/issues/83"
+sidebar:
+  label: 'Reverse Linked List II'
+  badge: 'Medium'
+---
+
+::::warning
+Could you do it in one pass?
+::::
+
+### Example 1:
+- Input: `head = [1,2,3,4,5], left = 2, right = 4`
+- Output: `[1,4,3,2,5]`
+
+### Example 2:
+- Input: `head = [5], left = 1, right = 1`
+- Output: `[5]`
+
+### Constraints:
+
+- The number of nodes in the list is n.
+- `1 <= n <= 500`
+- `-500 <= Node.val <= 500`
+- `1 <= left <= right <= n`
+
+## Solution
+
+```py
+# Definition for singly-linked list.
+# class ListNode:
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution:
+    def reverseBetween(
+        self, head: ListNode | None, left: int, right: int
+    ) -> ListNode | None:
+        cur, prev = head, None
+        while left > 1:
+            prev = cur
+            cur = cur.next
+            left, right = left - 1, right - 1
+
+        tail, con = cur, prev
+        while right:
+            third = cur.next
+            cur.next = prev
+            prev = cur
+            cur = third
+            right -= 1
+
+        if con:
+            con.next = prev
+        else:
+            head = prev
+
+        tail.next = cur
+        return head
+```