I psyched myself out...so I came up with a Python binary search solution 88% faster
Is this overkill?\n\n\ndef numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:\n\tif tomatoSlices % 2 == 1:\n\t\treturn []\n\n\tst = 0\n\ten 
Is this overkill?\n\n```\ndef numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:\n\tif tomatoSlices % 2 == 1:\n\t\treturn []\n\n\tst = 0\n\ten = cheeseSlices\n\n\twhile(st < en):\n\t\thalf = st + math.floor((en - st) / 2)\n\n\t\tsm = (half * 4) + ((cheeseSlices-half)*2)\n\t\tif sm == tomatoSlices:\n\t\t\treturn [half, cheeseSlices-half]\n\n\t\tif sm < tomatoSlices:\n\t\t\tst = half+1\n\t\telse:\n\t\t\ten = half\n\n\tif cheeseSlices * 4 == tomatoSlices:\n\t\treturn [cheeseSlices, 0]\n\n\treturn []\n```
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