✅ EASY SOLUTION PYTHON || BEATS 99% RUNTIME ✅
Here: \n1. count is the counter inside a function set to maintain the max min pattern ( if count is odd => find min; else => find max)\n2. final [] returns the 
Here: \n1. count is the counter inside a function set to maintain the max min pattern ( if count is odd => find min; else => find max)\n2. final [] returns the modified list with max min values as list values which replaces the nums\n3. repeating the process until we get only one element in nums\n# Code\n```\nclass Solution:\n    def minMaxGame(self, nums: List[int]) -> int:\n        def minmax(lis):\n            final = []\n            count = 1\n            for i in range(0,len(lis),2):\n                temp = lis[i:i+2]\n                if count%2 !=0:\n                    final.append(min(temp))\n                else:\n                    final.append(max(temp)) \n                count+=1\n            return final\n        while(len(nums)!=1):\n            nums = minmax(nums)\n        return nums[0]\n        \n```
['Python3']