✅ Python Easy Approach
\nclass Solution:\n    def minMaxGame(self, nums: List[int]) -> int:                \n        l=nums\n        while len(l)>1:\n            is_min=True     \n   
```\nclass Solution:\n    def minMaxGame(self, nums: List[int]) -> int:                \n        l=nums\n        while len(l)>1:\n            is_min=True     \n            tmp=[]\n            for i in range(0, len(l), 2):\n                if is_min:\n                    tmp.append(min(l[i:i+2]))\n                else:\n                    tmp.append(max(l[i:i+2]))\n                is_min=not is_min            \n            l=tmp            \n        return l[0]                        \n```\n\n**Time - O(nlogn)** - The input space is reduced by half after each iteration. \n**Space - O(n)**\n\n**Note**: The space can be reduced to `O(1)` if you use input array as the space to store the next list of elements. But I don\'t like the idea of modifying the input array.\n\n---\n\n***Please upvote if you find it useful***
['Python', 'Python3']