Double Listening Count

Problem Highlights

• 💡 Difficulty: Medium
• ⏰ Time to complete: 20-30 mins
• 🛠️ Topics: Linked Lists, Arithmetic Operations, Carry Forward

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• What does the problem ask for?
  • The problem asks to double the integer value represented by a linked list and return the head of the new list representing the doubled value.
• What should be returned?
  • The function should return the head of the modified linked list after doubling the integer value.

HAPPY CASE
Input: 
    - monthly_listeners1 = Node(1, Node(8, Node(9))) # 189
    - monthly_listeners2 = Node(9, Node(9, Node(9))) # 999
Output: 
    - 3 -> 7 -> 8
    - 1 -> 9 -> 9 -> 8
Explanation: 
    - 189 * 2 = 378
    - 999 * 2 = 1998

EDGE CASE
Input: 
    - monthly_listeners = Node(5, Node(0, Node(0))) # 500
Output: 
    - 1 -> 0 -> 0 -> 0
Explanation: 
    - 500 * 2 = 1000

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

For Linked List problems involving Arithmetic Operations and Carry Handling, we want to consider the following approaches:

• Traversal with Carry Handling: Traverse the linked list, double each node's value, and manage any carry that results from the doubling operation.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: We will traverse the linked list, double each node's value, handle any carry that results from the doubling, and create new nodes if there is a carry at the end of the list.

1) Initialize a pointer `current` to the head of the linked list.
2) Traverse the linked list:
    - Double the value of the current node.
    - Add any carry from the previous node's operation.
    - Update the node's value to be the result modulo 10.
    - Calculate the new carry to be passed to the next node.
3) If there is a carry left after traversing all nodes, create a new node with the carry value at the end of the list.
4) Return the head of the modified linked list.

⚠️ Common Mistakes

• Forgetting to handle cases where the carry is greater than 0 after the last node.
• Incorrectly managing pointers, leading to loss of nodes or incorrect traversal.

4: I-mplement

Implement the code to solve the algorithm.

class Node:
    def __init__(self, value, next=None):
        self.value = value
        self.next = next

def double_listeners(monthly_listeners):
    if not monthly_listeners:
        return None
    
    # Step 1: Traverse the list and double each node's value
    current = monthly_listeners
    carry = 0
    while current:
        doubled_value = current.value * 2 + carry
        current.value = doubled_value % 10
        carry = doubled_value // 10
        if current.next is None and carry > 0:
            # Add a new node for the carry at the end
            current.next = Node(carry)
            carry = 0
        current = current.next
    
    return monthly_listeners

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code's variables along the way.

• Example: Use the provided monthly_listeners1 and monthly_listeners2 linked lists to verify that the function correctly doubles the integer values represented by the linked lists.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

Assume N represents the length of the linked list.

• Time Complexity: O(N) because we traverse the entire linked list once.
• Space Complexity: O(1) because the algorithm uses a constant amount of extra space for pointers and carry.