Add Two Number — Day 15(Python)

Annamariya Tharayil

Oct 10, 2020·2 min read

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

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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