Leetcode_1276_Number of Burgers with No Waste of Ingredients

[lihe]

Number of Burgers with No Waste of Ingredients

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.

Example 4:

Input: tomatoSlices = 0, cheeseSlices = 0
Output: [0,0]

Example 5:

Input: tomatoSlices = 2, cheeseSlices = 1
Output: [0,1]

Constraints:

• 0 <= tomatoSlices <= 10^7
• 0 <= cheeseSlices <= 10^7

class Solution {
    public List<Integer> numOfBurgers(int tomatoSlices, int cheeseSlices) {
        List<Integer> result = new ArrayList<>();
        if(tomatoSlices > 4 * cheeseSlices || tomatoSlices < 2 * cheeseSlices || tomatoSlices % 2 != 0){
            return result;
        }

        int bigBurgers = tomatoSlices / 2 - cheeseSlices;
        int smallBurgers = 2 * cheeseSlices - tomatoSlices / 2;
        result.add(bigBurgers);
        result.add(smallBurgers);
        return result;
    }
}