📌 4 Liner || Constant Space
\tclass Solution {\n\tpublic:\n\t\tint minMaxGame(vector& nums) {\n\t\t\tfor(int n = nums.size();n > 1; n /= 2)\n\t\t\t\tfor (int i = 0; i < n / 2; i++) \n\t\t\
\tclass Solution {\n\tpublic:\n\t\tint minMaxGame(vector<int>& nums) {\n\t\t\tfor(int n = nums.size();n > 1; n /= 2)\n\t\t\t\tfor (int i = 0; i < n / 2; i++) \n\t\t\t\t\tnums[i] = (i%2 == 0) ? min(nums[2*i], nums[2*i + 1]) : max(nums[2*i], nums[2*i + 1]);\n\n\t\t\treturn nums[0];\n\t\t}\n\t};\n\t\n**Time Complexity :** *O(n)*\n**Space Complexity :** *O(1)*
['C']