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Solving Leetcode Interviews in Seconds with AI: Convert Binary Number in a Linked List to Integer

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2 min read

Introduction

In this blog post, we will explore how to solve the LeetCode problem "1290" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

Given head which is a reference node to a singly-linked list. The value of each node in the linked list is either 0 or 1. The linked list holds the binary representation of a number. Return the decimal value of the number in the linked list. The most significant bit is at the head of the linked list. Example 1: Input: head = [1,0,1] Output: 5 Explanation: (101) in base 2 = (5) in base 10 Example 2: Input: head = [0] Output: 0 Constraints: The Linked List is not empty. Number of nodes will not exceed 30. Each node's value is either 0 or 1.

Explanation

Here's the solution to convert a binary linked list to its decimal representation:

  • Approach: Iterate through the linked list, updating the decimal value at each step. For each node, left-shift the current decimal value by one bit (multiply by 2) and add the node's value. This effectively simulates the binary-to-decimal conversion process.
  • Optimal Solution: Initialize the decimal value to 0. Iterate through the linked list, updating the decimal_value as decimal_value = decimal_value * 2 + node.val.

  • Complexity:

    • Runtime: O(N), where N is the number of nodes in the linked list.
    • Storage: O(1), constant extra space.

Code

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

class Solution:
    def getDecimalValue(self, head: ListNode) -> int:
        decimal_value = 0
        current = head
        while current:
            decimal_value = (decimal_value * 2) + current.val
            current = current.next
        return decimal_value

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