# Add Two Number — Day 15(Python)

Today we will be looking at one of the frequently asked questions in a technical coding interview.

**2****. Add Two Numbers**

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:**

**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

- Initialize carry as 0 since we do not have a carry-forward number when we start our addition.
- While we have elements in both the linked-list add both numbers along with the carry-forward number.
- Consider the resulting answer unit digit as a node in the linked list and other digits as the carry-forward number.
- Repeat step 2 and 3 until one of the linked lists reach the end.
- If any of the linked-list is yet to be visited, add the remaining nodes with the carry-forward number.

`class LinklistAdder:`

def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:

new_node = ListNode(0)

main_node = new_node

carry = 0

while(l1 and l2):

addition = l1.val + l2.val + carry

new_node.next = ListNode(addition%10)

carry = addition//10

new_node = new_node.next

l1 = l1.next

l2 = l2.next

while l1:

addition = l1.val+carry

new_node.next = ListNode(addition%10)

carry = addition//10

new_node = new_node.next

l1 = l1.next

while l2:

addition = l2.val+carry

new_node.next = ListNode(addition%10)

carry = addition//10

new_node = new_node.next

l2 = l2.next

if carry:

new_node.next = ListNode(carry)

return main_node.next

**Time Complexity**

We are traversing through both the list at the same time and hence the time complexity is O(max(M, N)) where M, N is the length of both the LinkedList.

**Space Complexity**

We are creating a new LinkedList as an output that would take O(max(M, N)) where M, N is the length of both the LinkedList.

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