[C++, C] | Time Complexity - O(n), Space Complexity - O(1) Solution
If you like please upvote and any query you can ask in comment\n\t\n\tint minMaxGame(int* nums, int numsSize) {\n    \n\t\tint left = 0,\n\t\tright = numsSize -
**If you like please upvote and any query you can ask in comment**\n\t\n\tint minMaxGame(int* nums, int numsSize) {\n    \n\t\tint left = 0,\n\t\tright = numsSize - 1,\n\t\tindex = 0;\n\n\t\twhile (left <= right) {\n\n\t\t\tif (left == right) {\n\t\t\t\treturn nums[0];\n\t\t\t}\n\n\t\t\tif ((index & 1) == 0) {\n\t\t\t\t// even condition\n\t\t\t\tnums[index] = nums[index << 1] < nums[(index << 1) + 1] ? nums[index << 1] : nums[(index << 1) + 1];\n\t\t\t} else {\n\t\t\t\t// odd condition\n\t\t\t\tnums[index] = nums[index << 1] > nums[(index << 1) + 1] ? nums[index << 1] : nums[(index << 1) + 1];\n\t\t\t}\n\n\t\t\tif (index == right >> 1) {\n\t\t\t\tindex = 0;\n\t\t\t\tright >>= 1;\n\t\t\t} else {\n\t\t\t\tindex++;\n\t\t\t}\n\t\t}\n\n\t\treturn -1;\n\t}
['Two Pointers', 'C']