# Solving Leetcode Interviews in Seconds with AI: Min Max Game


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2293" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), 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 a 0-indexed integer array nums whose length is a power of 2. Apply the following algorithm on nums:  Let n be the length of nums. If n == 1, end the process. Otherwise, create a new 0-indexed integer array newNums of length n / 2. For every even index i where 0 <= i < n / 2, assign the value of newNums[i] as min(nums[2 * i], nums[2 * i + 1]). For every odd index i where 0 <= i < n / 2, assign the value of newNums[i] as max(nums[2 * i], nums[2 * i + 1]). Replace the array nums with newNums. Repeat the entire process starting from step 1.  Return the last number that remains in nums after applying the algorithm.   Example 1:   Input: nums = [1,3,5,2,4,8,2,2] Output: 1 Explanation: The following arrays are the results of applying the algorithm repeatedly. First: nums = [1,5,4,2] Second: nums = [1,4] Third: nums = [1] 1 is the last remaining number, so we return 1.  Example 2:  Input: nums = [3] Output: 3 Explanation: 3 is already the last remaining number, so we return 3.    Constraints:  1 <= nums.length <= 1024 1 <= nums[i] <= 109 nums.length is a power of 2.  

	# Explanation
	Here's the breakdown of the solution:

*   **Iterative Reduction:** The core idea is to simulate the algorithm step by step, iteratively reducing the size of the array `nums` until only one element remains.
*   **Min/Max Calculation:** In each iteration, create `newNums` by applying the min/max logic based on the index's parity.
*   **In-place Optimization:** Instead of creating a new array in each iteration, update `nums` in place. This optimizes space usage by reducing the number of arrays created.

*   **Runtime Complexity:** O(n), where n is the initial length of `nums`.
*   **Storage Complexity:** O(1). We are operating in place so the memory usage does not scale with the input size.

	
	# Code
	```python
	def min_max_game(nums: list[int]) -> int:
    """
    Applies the min-max game algorithm until a single number remains.

    Args:
        nums: A 0-indexed integer array whose length is a power of 2.

    Returns:
        The last number that remains in nums after applying the algorithm.
    """
    n = len(nums)

    while n > 1:
        new_nums = [0] * (n // 2)
        for i in range(n // 2):
            if i % 2 == 0:
                new_nums[i] = min(nums[2 * i], nums[2 * i + 1])
            else:
                new_nums[i] = max(nums[2 * i], nums[2 * i + 1])
        nums = new_nums
        n = len(nums)

    return nums[0]
	```