🚀Day 5. Add Number Linked Lists 🧠

The problem can be found at the following link: Question Link

💡 Problem Description:

You are given the head of two singly linked lists num1 and num2, which represent two non-negative integers. The digits are stored in reverse order, and each node contains a single digit. Return the head of the linked list representing the sum of these two numbers.

Note:

• There can be leading zeros in the input lists.

• The output list should not contain leading zeros.

🔍 Example Walkthrough:

Example 1:

Input: num1 = 4 -> 5, num2 = 3 -> 4 -> 5 Output: 3 -> 9 -> 0

image

Explanation: The given numbers are 45 and 345. Their sum is 390.

Example 2:

Input: num1 = 0 -> 0 -> 6 -> 3, num2 = 0 -> 7 Output: 7 -> 0

image

Explanation: The given numbers are 63 and 7. Their sum is 70.

Constraints:

• 1 <= size of both linked lists <= $10^6$

• 0 <= elements of both linked lists <= 9

🎯 My Approach:

To solve this problem, we use the following steps:

1. Reverse the Input Lists:

   • Since the input numbers are stored in reverse order, reverse both linked lists to process them in their natural order.

   • Use a helper function reverse() for this purpose.

2. Iterative Addition:

   • Traverse both linked lists simultaneously, summing their values along with a carry.

   • Create new nodes for the resultant linked list using the modulus of the sum (sum % 10), and update the carry as sum // 10.

3. Reverse the Resultant List:

   • After adding, reverse the resultant linked list to return the result in the required format.

4. Remove Leading Zeros:

   • If there are any leading zeros in the resultant list (other than a single zero), remove them for a clean output.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(max(n, m)), where n and m are the lengths of the input linked lists. Each list is traversed multiple times: once for reversing and once for summing.

• Expected Auxiliary Space Complexity: O(1), as no extra space proportional to the size of the input is used; only a constant amount of additional space is required.

📝 Solution Code

Code (C++)

class Solution {
public:
    Node* reverse(Node* head) {
        Node* prev = nullptr;
        Node* curr = head;
        while (curr) {
            Node* next = curr->next;
            curr->next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }

    Node* addTwoLists(Node* l1, Node* l2) {
        l1 = reverse(l1);
        l2 = reverse(l2);
        Node* dummy = new Node(0);
        Node* tail = dummy;
        int carry = 0;

        while (l1 || l2 || carry) {
            int sum = carry;
            if (l1) sum += l1->data, l1 = l1->next;
            if (l2) sum += l2->data, l2 = l2->next;
            carry = sum / 10;
            tail->next = new Node(sum % 10);
            tail = tail->next;
        }

        Node* res = reverse(dummy->next);
        delete dummy;
        while (res && res->data == 0 && res->next) {
            Node* temp = res;
            res = res->next;
            delete temp;
        }
        return res;
    }
};

Code (Java)

class Solution {
    Node reverse(Node head) {
        Node prev = null, curr = head;
        while (curr != null) {
            Node next = curr.next;
            curr.next = prev;
            prev = curr;
            curr = next;
        }
        return prev;
    }

    Node addTwoLists(Node l1, Node l2) {
        l1 = reverse(l1);
        l2 = reverse(l2);
        Node dummy = new Node(0);
        Node tail = dummy;
        int carry = 0;

        while (l1 != null || l2 != null || carry != 0) {
            int sum = carry;
            if (l1 != null) {
                sum += l1.data;
                l1 = l1.next;
            }
            if (l2 != null) {
                sum += l2.data;
                l2 = l2.next;
            }
            carry = sum / 10;
            tail.next = new Node(sum % 10);
            tail = tail.next;
        }

        Node res = reverse(dummy.next);
        while (res != null && res.data == 0 && res.next != null) {
            res = res.next;
        }
        return res;
    }
}

Code (Python)

class Solution:
    def reverse(self, head):
        prev, curr = None, head
        while curr:
            next_node = curr.next
            curr.next = prev
            prev = curr
            curr = next_node
        return prev

    def addTwoLists(self, l1, l2):
        l1 = self.reverse(l1)
        l2 = self.reverse(l2)
        dummy = Node(0)
        tail = dummy
        carry = 0

        while l1 or l2 or carry:
            summ = carry
            if l1:
                summ += l1.data
                l1 = l1.next
            if l2:
                summ += l2.data
                l2 = l2.next
            carry = summ // 10
            tail.next = Node(summ % 10)
            tail = tail.next

        res = self.reverse(dummy.next)
        while res and res.data == 0 and res.next:
            res = res.next
        return res

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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