[LeetCode] 1276. Number of Burgers with No Waste of Ingredients

[grandyang]

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

这道题是关于做汉堡的问题，说是给了一些西红柿切片和芝士切片，让做两种汉堡，一种是巨无霸汉堡，需要4片西红柿切片和1片芝士切片，另一种是小汉堡，需要2片西红柿和1片芝士切片，问正好用完所有的切片，能分别做出两种汉堡多少个。这道题让博主想到了《分手厨房》这款小游戏，咚咚咚的切西红柿的声音在脑边回想起来，盘子扔的飞起。这里说了两种汉堡都只需要一片芝士，那么芝士的总数一定就是两个汉堡的个数，所以最简单的办法就是遍历和为 cheeseSlices 的两个数字的所有组合，然后再去计算其需要的西红柿切片个数是否正好等于给定的 tomatoSlices 就可以了，感觉应该算一道 Easy 的题目，参见代码如下：

解法一：

class Solution {
public:
    vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) {
        for (int i = 0; i <= cheeseSlices; ++i) {
            if (i * 4 + (cheeseSlices - i) * 2 == tomatoSlices) {
                return {i, cheeseSlices - i};
            }
        }
        return {};
    }
};

这道题其实是一道二元一次方程求解的题目，基本算是初中数学的知识，设巨无霸汉堡有x个，小汉堡有y个，西红柿切片个数为t，芝士切片个数为c，则可写出以下两个方程:

4x + 2y = t
x + y = c

求解可得：
x = t / 2 - c
y = 2c - t / 2

由于x和y的解必须是非负数，所有下列两个不等式必须同时成立：
t / 2 - c >= 0
2c - t / 2 >= 0

化简可得 t >= 2c && t <= 4c，同时，因为两种汉堡都是需要偶数个西红柿切片，所以t必须是偶数，只要同时满足这个三个条件的t和c，才能得到x和y的正确解，所以其实这要一行代码就可以了，博主加了一行变量名称化简的操作，当三个条件满足的时候，直接按上面的等式求出x和y返回，否则返回空数组即可，参见代码如下：

解法二：

class Solution {
public:
    vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) {
        int t = tomatoSlices, c = cheeseSlices;
        return (t % 2 == 0 && t >= 2 * c && t <= 4 * c) ? vector<int>{t / 2 - c, 2 * c - t / 2} : vector<int>();
    }
};

Github 同步地址:

#1276

参考资料：

https://leetcode.com/problems/number-of-burgers-with-no-waste-of-ingredients/

https://leetcode.com/problems/number-of-burgers-with-no-waste-of-ingredients/discuss/441475/The-only-realistic-solution

https://leetcode.com/problems/number-of-burgers-with-no-waste-of-ingredients/discuss/441321/JavaC%2B%2BPython-Chickens-and-Rabbits

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