diff --git a/apps/docs/content/(array)/48-rotate-image.mdx b/apps/docs/content/(array)/48-rotate-image.mdx
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+---
+title: '48. Rotate Image'
+description: You are given an n x n 2D matrix representing an image, rotate the image by 90 degrees (clockwise)
+sidebar:
+  label: 'Rotate Image'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Matrix Index Math</Badge>
+
+### Example 1:
+- Input: `matrix = [[1,2,3],[4,5,6],[7,8,9]]`
+- Output: `[[7,4,1],[8,5,2],[9,6,3]]`
+
+### Example 2:
+- Input: `matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]`
+- Output: `[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]`
+
+### Constraints:
+
+- `n == matrix.length == matrix[i].length`
+- `1 <= n <= 20`
+- `-1000 <= matrix[i][j] <= 1000`
+
+## Solution
+
+```py
+class Solution:
+    def rotate(self, matrix: List[List[int]]) -> None:
+        n = len(matrix)
+        for r in range(n):
+            for c in range(r + 1, n):
+                matrix[r][c], matrix[c][r] = matrix[c][r], matrix[r][c]
+
+        for r in range(n):
+            matrix[r].reverse()
+```
diff --git a/apps/docs/content/(array)/54-spiral-matrix.mdx b/apps/docs/content/(array)/54-spiral-matrix.mdx
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+---
+title: '54. Spiral Matrix'
+description: Given an m x n matrix, return all elements of the matrix in spiral order
+sidebar:
+  label: 'Spiral Matrix'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Matrix Index Math</Badge>
+
+### Example 1:
+- Input: `matrix = [[1,2,3],[4,5,6],[7,8,9]]`
+- Output: `[1,2,3,6,9,8,7,4,5]`
+
+### Example 2:
+- Input: `matrix = [[1,2,3,4],[5,6,7,8],[9,10,11,12]]`
+- Output: `[1,2,3,4,8,12,11,10,9,5,6,7]`
+
+### Constraints:
+
+- `m == matrix.length`
+- `n == matrix[i].length`
+- `1 <= m, n <= 10`
+- `-100 <= matrix[i][j] <= 100`
+
+## Solution
+
+```py
+class Solution:
+    def spiralOrder(self, matrix: List[List[int]]) -> List[int]:
+        res = []
+        top, bottom = 0, len(matrix) - 1
+        left, right = 0, len(matrix[0]) - 1
+
+        while top <= bottom and left <= right:
+            # 1. Top Row
+            for c in range(left, right + 1):
+                res.append(matrix[top][c])
+            top += 1
+
+            # 2. Right Column
+            for r in range(top, bottom + 1):
+                res.append(matrix[r][right])
+            right -= 1
+
+            # 3. Bottom Row
+            if top <= bottom:
+                for c in range(right, left - 1, -1):
+                    res.append(matrix[bottom][c])
+                bottom -= 1
+
+            # 4. Left Column
+            if left <= right:
+                for r in range(bottom, top - 1, -1):
+                    res.append(matrix[r][left])
+                left += 1
+
+        return res
+```
diff --git a/apps/docs/content/(linked-list)/2-add-two-numbers.mdx b/apps/docs/content/(linked-list)/2-add-two-numbers.mdx
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+---
+title: '2. Add Two Numbers'
+description: You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each of their nodes contains a single digit. Add the two numbers and return the sum as a linked list
+sidebar:
+  label: 'Add Two Numbers'
+  badge: 'Medium'
+---
+
+<Badge variant="accent">Hybrid Structures</Badge>
+
+### Example 1:
+- Input: `l1 = [2,4,3], l2 = [5,6,4]`
+- Output: `[7,0,8]`
+- Explanation: `342 + 465 = 807`.
+
+### Example 2:
+- Input: `l1 = [0], l2 = [0]`
+- Output: `[0]`
+
+### Example 3:
+- Input: `l1 = [9,9,9,9,9,9,9], l2 = [9,9,9,9]`
+- Output: `[8,9,9,9,0,0,0,1]`
+
+### Constraints:
+
+- The number of nodes in each linked list is in the range [1, 100].
+- `0 <= Node.val <= 9`
+- It is guaranteed that the list represents a number that does not have leading zeros.
+
+## 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 addTwoNumbers(
+        self, l1: Optional[ListNode], l2: Optional[ListNode]
+    ) -> Optional[ListNode]:
+
+        dummy = ListNode()
+        carry = 0
+        curr = dummy
+        while l1 or l2 or carry:
+            s = (l1.val if l1 else 0) + (l2.val if l2 else 0) + carry
+            carry, val = divmod(s, 10)
+
+            curr.next = ListNode(val)
+            curr = curr.next
+
+            l1 = l1.next if l1 else None
+            l2 = l2.next if l2 else None
+
+        return dummy.next
+```