Introduction

In this blog post, we will explore how to solve the LeetCode problem "1860" 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 two integers memory1 and memory2 representing the available memory in bits on two memory sticks. There is currently a faulty program running that consumes an increasing amount of memory every second. At the ith second (starting from 1), i bits of memory are allocated to the stick with more available memory (or from the first memory stick if both have the same available memory). If neither stick has at least i bits of available memory, the program crashes. Return an array containing [crashTime, memory1crash, memory2crash], where crashTime is the time (in seconds) when the program crashed and memory1crash and memory2crash are the available bits of memory in the first and second sticks respectively. Example 1: Input: memory1 = 2, memory2 = 2 Output: [3,1,0] Explanation: The memory is allocated as follows: - At the 1st second, 1 bit of memory is allocated to stick 1. The first stick now has 1 bit of available memory. - At the 2nd second, 2 bits of memory are allocated to stick 2. The second stick now has 0 bits of available memory. - At the 3rd second, the program crashes. The sticks have 1 and 0 bits available respectively. Example 2: Input: memory1 = 8, memory2 = 11 Output: [6,0,4] Explanation: The memory is allocated as follows: - At the 1st second, 1 bit of memory is allocated to stick 2. The second stick now has 10 bit of available memory. - At the 2nd second, 2 bits of memory are allocated to stick 2. The second stick now has 8 bits of available memory. - At the 3rd second, 3 bits of memory are allocated to stick 1. The first stick now has 5 bits of available memory. - At the 4th second, 4 bits of memory are allocated to stick 2. The second stick now has 4 bits of available memory. - At the 5th second, 5 bits of memory are allocated to stick 1. The first stick now has 0 bits of available memory. - At the 6th second, the program crashes. The sticks have 0 and 4 bits available respectively. Constraints: 0 <= memory1, memory2 <= 231 - 1

Explanation

• The solution simulates the memory allocation process step by step until a crash occurs. In each step, it checks which memory stick has more available memory and allocates the required amount.
  • If neither stick has enough memory, the loop breaks, and the crash time and remaining memory are recorded.
  • The solution uses a while loop and simple if-else statements to implement the logic efficiently.

• Runtime Complexity: O(N), where N is the crash time. In the worst case, where memory1 and memory2 are large enough, the loop iterates until i becomes significantly large. In the best case, the complexity will be O(1) where memory1 and memory2 are 0 and crash happens instantly. Storage Complexity: O(1) because only a fixed number of variables are used.

Code

    def memory_allocation(memory1: int, memory2: int) -> list[int]:
    """
    Simulates a faulty program's memory allocation and returns the crash time and memory state.

    Args:
        memory1: Initial memory available on stick 1.
        memory2: Initial memory available on stick 2.

    Returns:
        A list containing [crashTime, memory1crash, memory2crash].
    """
    i = 1
    while True:
        if memory1 >= memory2:
            if memory1 >= i:
                memory1 -= i
            else:
                return [i, memory1, memory2]
        else:
            if memory2 >= i:
                memory2 -= i
            else:
                return [i, memory1, memory2]
        i += 1