[Java/Python 3] Solve a linear equations group w/ brief explanation and analysis.
According the description of the problem, we have the following:\n\nJumbo Burger: 4 tomato slices and 1 cheese slice.\nSmall Burger: 2 Tomato slices and 1 chees
According the description of the problem, we have the following:\n```\nJumbo Burger: 4 tomato slices and 1 cheese slice.\nSmall Burger: 2 Tomato slices and 1 cheese slice.\n```\nTherefore:\n```\n4 * Jumbo Burger + 2 * Small Burger = tomatoSlices\nJumbo Burger + Small Burger = cheeseSlices\n```\nLet `x` and `y` indicate `jumbo` and `small` Burger, respectively, then\n```\n4x + 2y = tomatoSlices\nx + y = cheeseSlices\n```\n=>\n```\n2x = tomatoSlices - 2 * cheeseSlices\n```\n=>\n```\nx = (tomatoSlices - 2 * cheeseSlices) / 2\ny = cheeseSlices - x\n```\n\n----\n\n```java\n    public List<Integer> numOfBurgers(int tomatoSlices, int cheeseSlices) {\n        int twoX = tomatoSlices - 2 * cheeseSlices, x = twoX / 2, y = cheeseSlices - x;\n        return twoX >= 0 && twoX % 2 == 0 && y >= 0 ? Arrays.asList(x, y) : Arrays.asList();        \n    }\n```\n```python\n    def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:\n        two_x = tomatoSlices - 2 * cheeseSlices\n        x = two_x // 2\n        y = cheeseSlices - x\n        return [x, y] if two_x >= 0 and not two_x % 2 and y >= 0 else []\n```\n**Analysis**\n\nTime & space: O(1).
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