Introduction

In this blog post, we will explore how to solve the LeetCode problem "2154" 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

You are given an array of integers nums. You are also given an integer original which is the first number that needs to be searched for in nums. You then do the following steps: If original is found in nums, multiply it by two (i.e., set original = 2 * original). Otherwise, stop the process. Repeat this process with the new number as long as you keep finding the number. Return the final value of original. Example 1: Input: nums = [5,3,6,1,12], original = 3 Output: 24 Explanation: - 3 is found in nums. 3 is multiplied by 2 to obtain 6. - 6 is found in nums. 6 is multiplied by 2 to obtain 12. - 12 is found in nums. 12 is multiplied by 2 to obtain 24. - 24 is not found in nums. Thus, 24 is returned. Example 2: Input: nums = [2,7,9], original = 4 Output: 4 Explanation: - 4 is not found in nums. Thus, 4 is returned. Constraints: 1 <= nums.length <= 1000 1 <= nums[i], original <= 1000

Explanation

Here's an efficient solution to the problem:

• Use a Set for Efficient Lookup: Store the numbers in nums in a set. This allows for O(1) average time complexity for checking if a number exists in nums.
• Iterative Multiplication: Repeatedly multiply original by 2 as long as the new value is present in the set.

• Return Final Value: Once original is no longer found, return its final value.

• Runtime Complexity: O(n), where n is the length of the nums array (due to set creation). In the best case where the loop does not execute, the complexity would be O(1). The loop execution depends on original and the values of nums.

• Storage Complexity: O(n), where n is the length of the nums array (due to the set).

Code

    def find_final_value(nums, original):
    """
    Finds the final value of original after repeatedly multiplying by 2 if found in nums.

    Args:
        nums: A list of integers.
        original: The starting integer value.

    Returns:
        The final value of original.
    """
    nums_set = set(nums)  # Convert nums to a set for efficient lookup

    while original in nums_set:
        original *= 2

    return original