# Add Two Number — Day 15(Python)

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

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 = , l2 = Output: `

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:

Solution

1. Initialize carry as 0 since we do not have a carry-forward number when we start our addition.
2. While we have elements in both the linked-list add both numbers along with the carry-forward number.
3. Consider the resulting answer unit digit as a node in the linked list and other digits as the carry-forward number.
4. Repeat step 2 and 3 until one of the linked lists reach the end.
5. 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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