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+---
+id: add-two-numbers
+title: Two Sum Problem (LeetCode)
+sidebar_label: 0002 - Add Two Numbers
+tags:
+  - Linked List
+  - Math
+  - Recursion
+description: "This is a solution to the Add Two Numbers problem on LeetCode."
+---
+## Problem Description
+
+| Problem Statement                                             | Solution Link                                                                                                                                                             | LeetCode Profile                                    |
+| :------------------------------------------------------------ | :------------------------------------------------------------------------------------------------------------------------------------------------------------------------ | :-------------------------------------------------- |
+| [Add Two Numbers on LeetCode](https://leetcode.com/problems/add-two-numbers/) | [Add Two Numbers Solution on LeetCode](https://leetcode.com/problems/add-two-numbers/solutions/5234194/solution/) | [Amruta Jayanti](https://leetcode.com/u/user7669cY/)|
+
+
+## Problem 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.
+
+You may assume the two numbers do not contain any leading zero, except the number 0 itself.
+
+ 
+
+**Example 1:**
+
+```plaintext
+Input: l1 = [2,4,3], l2 = [5,6,4]
+Output: [7,0,8]
+Explanation: 342 + 465 = 807.
+```
+
+**Example 2:**
+
+```plaintext
+Input: l1 = [0], l2 = [0]
+Output: [0]
+```
+
+**Example 3:**
+
+```plaintext
+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 to the problem
+
+### Intuition and Approach
+This is a simple problem. It can be done by maintaining two pointers , each for each linked list. Add the values store them and then move to next node.
+
+#### Implementation
+```python
+# Definition for singly-linked list.
+# class ListNode(object):
+#     def __init__(self, val=0, next=None):
+#         self.val = val
+#         self.next = next
+class Solution(object):
+    def addTwoNumbers(self, l1, l2):
+        """
+        :type l1: ListNode
+        :type l2: ListNode
+        :rtype: ListNode
+        """
+        carry = 0
+        dummy = ListNode()
+        current = dummy
+
+        while l1 or l2 or carry:
+            # Extract values from the current nodes
+            val1 = l1.val if l1 else 0
+            val2 = l2.val if l2 else 0
+
+            # Calculate the sum and carry
+            total_sum = val1 + val2 + carry
+            carry = total_sum // 10
+            current.next = ListNode(total_sum % 10)
+
+            # Move to the next nodes
+            if l1:
+                l1 = l1.next
+            if l2:
+                l2 = l2.next
+
+            # Move to the next result node
+            current = current.next
+
+        return dummy.next
+```
+Above is the implementation in Python. Here total_sum stores the value and adds to the dummy. Variable carry is used to handle the carry bits.
+
+#### Complexity Analysis:
+- Time Complexity : $$O(max(n,m))$$ Here, n,m are the lengths of the input linked lists
+- Space Complexity : $$O(max(n,m))$$
+
+#### Codes in different languages:
+`CPP`:
+```cpp
+class Solution {
+public:
+    ListNode* addTwoNumbers(ListNode* l1, ListNode* l2) {
+        ListNode* dummyHead = new ListNode(0);
+        ListNode* tail = dummyHead;
+        int carry = 0;
+
+        while (l1 != nullptr || l2 != nullptr || carry != 0) {
+            int digit1 = (l1 != nullptr) ? l1->val : 0;
+            int digit2 = (l2 != nullptr) ? l2->val : 0;
+
+            int sum = digit1 + digit2 + carry;
+            int digit = sum % 10;
+            carry = sum / 10;
+
+            ListNode* newNode = new ListNode(digit);
+            tail->next = newNode;
+            tail = tail->next;
+
+            l1 = (l1 != nullptr) ? l1->next : nullptr;
+            l2 = (l2 != nullptr) ? l2->next : nullptr;
+        }
+
+        ListNode* result = dummyHead->next;
+        delete dummyHead;
+        return result;
+    }
+};
+```
+
+`Java`:
+```java
+class Solution {
+    public ListNode addTwoNumbers(ListNode l1, ListNode l2) {
+        ListNode dummyHead = new ListNode(0);
+        ListNode tail = dummyHead;
+        int carry = 0;
+
+        while (l1 != null || l2 != null || carry != 0) {
+            int digit1 = (l1 != null) ? l1.val : 0;
+            int digit2 = (l2 != null) ? l2.val : 0;
+
+            int sum = digit1 + digit2 + carry;
+            int digit = sum % 10;
+            carry = sum / 10;
+
+            ListNode newNode = new ListNode(digit);
+            tail.next = newNode;
+            tail = tail.next;
+
+            l1 = (l1 != null) ? l1.next : null;
+            l2 = (l2 != null) ? l2.next : null;
+        }
+
+        ListNode result = dummyHead.next;
+        dummyHead.next = null;
+        return result;
+    }
+}
+```
+
+
+