2293. Min Max Game — Daily Problem

Although it's a simple problem, it involves a recursive idea.

The approach is to continuously convert the 1D list into a 2D list for processing (this avoids index confusion caused by in-place changes), then use a flag counter to compare maximum and minimum values.

At the end, return the only remaining number — though we know it must be nums[0] by this point, we still use nums[0] to avoid warnings.

class Solution:
    def minMaxGame(self, nums: List[int]) -> int:
        flag = 0
        while len(nums) > 1:
            nums = [nums[i:i + 2] for i in range(0, len(nums), 2)]
            # split into 2D
            for i in range(len(nums)):
                if flag % 2 == 0:
                    nums[i] = min(nums[i])
                    flag += 1
                else:
                    nums[i] = max(nums[i])
                    flag += 1
        return nums[0]

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longsizhuo贡献 4 次 · 最近 2026/05/06

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2293. Min Max Game — Daily Problem

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