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Marc Peejay V. Viernes

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📝 Mastering Linked Lists Arithmetic: Solving the "Add Two Numbers" Problem 📝

Welcome to another adventure in the realm of data structures and algorithms! Today, we'll delve into the fascinating "Add Two Numbers" problem, a medium-level challenge that tests our ability to perform arithmetic operations on linked lists efficiently.

Problem Statement

You are given two non-empty linked lists representing two non-negative integers. The digits are stored in reverse order, and each node contains a single digit. The task is to add the two numbers and return the sum as a linked list.

Examples

Let's dive into a few examples to understand the problem better:

1. Example 1:

• Input: `l1 = [2,4,3]`, `l2 = [5,6,4]`
• Output: `[7,0,8]`
• Explanation: 342 + 465 = 807.
2. Example 2:

• Input: `l1 = [0]`, `l2 = [0]`
• Output: `[0]`
3. 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 Approach

To solve this problem, we'll traverse both linked lists simultaneously, performing digit-by-digit addition and propagating any carry that occurs. Here's our approach:

``````# Definition for singly-linked list.
class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next

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

carry = 0
while l1 or l2 or carry:
value1 = l1.val if l1 else 0
value2 = l2.val if l2 else 0

# Calculate the new digit.
val = value1 + value2 + carry
carry = val // 10
val = val % 10
currentNode.next = ListNode(val)

# Update the pointers.
currentNode = currentNode.next
l1 = l1.next if l1 else None
l2 = l2.next if l2 else None

``````

Algorithm

Just like how you would sum two numbers on a piece of paper, we begin by summing the least-significant digits, which is the head of l1 and l2. Since each digit is in the range of 0 to 9, summing two digits may result in an overflow. For example, 5 + 7 = 12. In this case, we set the current digit to 2 and bring over the carry of 1 to the next iteration. The carry must be either 0 or 1 because the largest possible sum of two digits (including the carry) is 9 + 9 + 1 = 19.

The pseudocode is as follows:

• Initialize current node to sum head node of the returning list.
• Initialize carry to 0.
• Loop through lists l1 and l2 until you reach both ends and carry is 0.
• Set x to node l1's value. If l1 has reached the end, set x to 0.
• Set y to node l2's value. If l2 has reached the end, set y to 0.
• Set sum = x + y + carry.
• Update carry = sum // 10.
• Create a new node with the digit value of sum % 10 and set it to current node's next, then advance current node to next.
• Advance both l1 and l2.
• Return dummy head's next node.

Complexity Analysis

• Time complexity: O(max(m,n)), where m and n represent the lengths of l1 and l2 respectively. The algorithm above iterates at most max(m,n) times.
• Space complexity: O(1). The length of the new list is at most max(m,n) + 1. However, we don't count the answer as part of the space complexity.

Conclusion

The "Add Two Numbers" problem challenges us to leverage linked list manipulation and basic arithmetic operations to compute the sum of large numbers efficiently. By implementing a systematic traversal strategy and handling carry propagation effectively, we can achieve accurate results for various test cases. This problem highlights the importance of algorithmic thinking and problem-solving skills in navigating complex challenges.

For further exploration and practice, visit the "Add Two Numbers" problem on LeetCode. Happy coding! 🚀