How do you solve Double a Number Represented as a Linked List on LeetCode?

Double a Number Represented as a Linked List (LeetCode #2816) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Understanding how to manage carry-over is crucial when dealing with digit manipulation.

What is the brute force solution for Double a Number Represented as a Linked List?

The brute force approach for Double a Number Represented as a Linked List is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves converting the linked list into a number, doubling that number, and then converting it back into a linked list. This is straightforward but inefficient for large numbers. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Double a Number Represented as a Linked List?

Double a Number Represented as a Linked List uses the following concepts: Linked List, Math, Stack. The recommended patterns to study are: Stack, Two Pointers.

What are the interview tips for Double a Number Represented as a Linked List?

Always clarify the constraints and edge cases before diving into coding. Think about how you can optimize your solution after writing the brute-force version. Practice explaining your thought process clearly while coding, as communication is key in interviews.

What are common mistakes when solving Double a Number Represented as a Linked List?

Forgetting to handle the carry when the last digit results in a value greater than 9. Not considering the edge case where the final carry results in an additional node.

#2816

Double a Number Represented as a Linked List

Medium

Linked ListMathStackStackTwo Pointers

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal approach involves traversing the linked list from the least significant digit to the most significant digit, doubling each value and managing carry-over directly. This avoids the overhead of converting to and from a number.

⚙️

Algorithm

4 steps

1Step 1: Initialize a carry variable to 0.
2Step 2: Traverse the linked list from the end to the start, doubling each node's value and adding the carry.
3Step 3: If the doubled value is greater than 9, set the current node's value to (doubled value % 10) and update the carry.
4Step 4: If there's a carry left after processing all nodes, create a new node for the carry.

solution.py22 lines

1class ListNode:
2    def __init__(self, val=0, next=None):
3        self.val = val
4        self.next = next
5
6def doubleLinkedList(head):
7    carry = 0
8    current = head
9    stack = []
10    while current:
11        stack.append(current)
12        current = current.next
13    while stack:
14        node = stack.pop()
15        doubled_value = node.val * 2 + carry
16        node.val = doubled_value % 10
17        carry = doubled_value // 10
18    if carry:
19        new_head = ListNode(carry)
20        new_head.next = head
21        return new_head
22    return head

ℹ

Complexity note: The time complexity is O(n) because we traverse the linked list once to push nodes onto the stack and then again to process them. The space complexity is O(n) due to the stack storing the nodes.

1Understanding how to manage carry-over is crucial when dealing with digit manipulation.
2Using a stack allows us to reverse the order of processing, which is essential for linked lists representing numbers.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.