1276. 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> res = new ArrayList();
        if(tomatoSlices == 0 && cheeseSlices == 0){
            res.add(0);
            res.add(0);
             return res;
        }
       
        int half = cheeseSlices / 2;
        //int[][] arr = new int[half + 1][2];
        for(int i = 0; i <= half; i++){
            int t1 =  i;
            int t2 = cheeseSlices - i;
            if(t1 * 4 + t2 * 2 == tomatoSlices){
                res.add(t1);
                res.add(t2);
            }
            else if(t1 * 2 + t2 * 4 == tomatoSlices){
                res.add(t2);
                res.add(t1);
            }
        }

        return res;
    }
}

CNM, O（n）都给TLE了

看了下解答，竟然是鸡兔同笼问题，一想确实有道理

class Solution {
    public List<Integer> numOfBurgers(int t, int c) {
        List<Integer> res = new ArrayList();
        if(t % 2 == 0 && c * 2 <= t && t <= c * 4){
            res.add(t/2 - c);
            res.add(c * 2 - t / 2);
        }
        else return res;
        return res;
        
    }
}

 https://leetcode.com/problems/number-of-burgers-with-no-waste-of-ingredients/discuss/441342/JavaPython-3-Solve-a-linear-equations-group-w-brief-explanation-and-analysis.