# 1276. Number of Burgers with No Waste of Ingredients

## Description

Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:

• Jumbo Burger: 4 tomato slices and 1 cheese slice.
• Small Burger: 2 Tomato slices and 1 cheese slice.

Return [total_jumbo, total_small] so that the number of remaining tomatoSlices equal to 0 and the number of remaining cheeseSlices equal to 0. If it is not possible to make the remaining tomatoSlices and cheeseSlices equal to 0 return [].

Example 1:

Input: tomatoSlices = 16, cheeseSlices = 7
Output: [1,6]
Explantion: To make one jumbo burger and 6 small burgers we need 4*1 + 2*6 = 16 tomato and 1 + 6 = 7 cheese.
There will be no remaining ingredients.


Example 2:

Input: tomatoSlices = 17, cheeseSlices = 4
Output: []
Explantion: There will be no way to use all ingredients to make small and jumbo burgers.


Example 3:

Input: tomatoSlices = 4, cheeseSlices = 17
Output: []
Explantion: Making 1 jumbo burger there will be 16 cheese remaining and making 2 small burgers there will be 15 cheese remaining.


Constraints:

• 0 <= tomatoSlices, cheeseSlices <= 107

## Solutions

Solution 1: Mathematics

We set the number of Jumbo Burgers as $x$ and the number of Small Burgers as $y$, then we have:

\begin{aligned} 4x + 2y &= tomatoSlices \\ x + y &= cheeseSlices \end{aligned}

Transforming the above two equations, we can get:

\begin{aligned} y = (4 \times cheeseSlices - tomatoSlices) / 2 \\ x = cheeseSlices - y \end{aligned}

Where $x$ and $y$ must be non-negative integers.

The time complexity is $O(1)$, and the space complexity is $O(1)$.

• class Solution {
public List<Integer> numOfBurgers(int tomatoSlices, int cheeseSlices) {
int k = 4 * cheeseSlices - tomatoSlices;
int y = k / 2;
int x = cheeseSlices - y;
return k % 2 != 0 || y < 0 || x < 0 ? List.of() : List.of(x, y);
}
}

• class Solution {
public:
vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) {
int k = 4 * cheeseSlices - tomatoSlices;
int y = k / 2;
int x = cheeseSlices - y;
return k % 2 || x < 0 || y < 0 ? vector<int>{} : vector<int>{x, y};
}
};

• class Solution:
def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:
k = 4 * cheeseSlices - tomatoSlices
y = k // 2
x = cheeseSlices - y
return [] if k % 2 or y < 0 or x < 0 else [x, y]


• func numOfBurgers(tomatoSlices int, cheeseSlices int) []int {
k := 4*cheeseSlices - tomatoSlices
y := k / 2
x := cheeseSlices - y
if k%2 != 0 || x < 0 || y < 0 {
return []int{}
}
return []int{x, y}
}

• function numOfBurgers(tomatoSlices: number, cheeseSlices: number): number[] {
const k = 4 * cheeseSlices - tomatoSlices;
const y = k >> 1;
const x = cheeseSlices - y;
return k % 2 || y < 0 || x < 0 ? [] : [x, y];
}


• impl Solution {
pub fn num_of_burgers(tomato_slices: i32, cheese_slices: i32) -> Vec<i32> {
let k = 4 * cheese_slices - tomato_slices;
let y = k / 2;
let x = cheese_slices - y;
if k % 2 != 0 || y < 0 || x < 0 {
Vec::new()
} else {
vec![x, y]
}
}
}