Introduction

In this blog post, we will explore how to solve the LeetCode problem "390" 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 have a list arr of all integers in the range [1, n] sorted in a strictly increasing order. Apply the following algorithm on arr: Starting from left to right, remove the first number and every other number afterward until you reach the end of the list. Repeat the previous step again, but this time from right to left, remove the rightmost number and every other number from the remaining numbers. Keep repeating the steps again, alternating left to right and right to left, until a single number remains. Given the integer n, return the last number that remains in arr. Example 1: Input: n = 9 Output: 6 Explanation: arr = [1, 2, 3, 4, 5, 6, 7, 8, 9] arr = [2, 4, 6, 8] arr = [2, 6] arr = [6] Example 2: Input: n = 1 Output: 1 Constraints: 1 <= n <= 109

Explanation

Here's a breakdown of the solution and the corresponding Python code:

• High-Level Approach:

  • Simulate the process efficiently without explicitly creating and modifying the array at each step.
  • Track the start (the smallest number in the current range), step (the increment between consecutive numbers), length (number of elements in current range), and direction (left to right or right to left).
  • Update these variables based on whether the length is even or odd and the direction of removal.

• Complexity Analysis:

  • Runtime Complexity: O(log n) - The number of iterations is proportional to how many times n can be divided by 2.
  • Storage Complexity: O(1) - Constant extra space is used.

Code

    def lastRemaining(n: int) -> int:
    start = 1
    step = 1
    length = n
    direction = True  # True for left to right, False for right to left

    while length > 1:
        if direction or length % 2 == 1:
            start = start + step
        length = length // 2
        step = step * 2
        direction = not direction

    return start