What is the main pattern to solve Number of Burgers with No Waste of Ingredients?

The core pattern is a math-driven system of linear equations that ensures all tomato and cheese slices are used with integer counts of burgers.

Can tomatoSlices be odd for a valid solution?

No, because jumbo and small burgers require an even number of tomato slices combined; an odd total makes zero-waste impossible.

What should I return if no combination of burgers uses all ingredients?

Return an empty array, as there is no way to satisfy zero leftover constraints with the given tomatoSlices and cheeseSlices.

Does this problem require iteration over possible burger counts?

No, direct algebraic computation suffices, making the solution O(1) time and avoiding unnecessary loops.

How do I verify the computed jumbo and small counts?

Ensure both are non-negative integers and that 4*jumbo + 2*small equals tomatoSlices and jumbo + small equals cheeseSlices.

#1276

Medium

auto_awesome数学·driven

LeetCode 题解工作台

不浪费原料的汉堡制作方案

圣诞活动预热开始啦，汉堡店推出了全新的汉堡套餐。为了避免浪费原料，请你帮他们制定合适的制作计划。给你两个整数 tomatoSlices 和 cheeseSlices ，分别表示番茄片和奶酪片的数目。不同汉堡的原料搭配如下：巨无霸汉堡： 4 片番茄和 1 片奶酪小皇堡： 2 片番茄和 1 片奶酪…

数学

题目描述

圣诞活动预热开始啦，汉堡店推出了全新的汉堡套餐。为了避免浪费原料，请你帮他们制定合适的制作计划。

给你两个整数 tomatoSlices 和 cheeseSlices，分别表示番茄片和奶酪片的数目。不同汉堡的原料搭配如下：

巨无霸汉堡：4 片番茄和 1 片奶酪
小皇堡：2 片番茄和 1 片奶酪

请你以 [total_jumbo, total_small]（[巨无霸汉堡总数，小皇堡总数]）的格式返回恰当的制作方案，使得剩下的番茄片 tomatoSlices 和奶酪片 cheeseSlices 的数量都是 0。

如果无法使剩下的番茄片 tomatoSlices 和奶酪片 cheeseSlices 的数量为 0，就请返回 []。

示例 1：

输入：tomatoSlices = 16, cheeseSlices = 7
输出：[1,6]
解释：制作 1 个巨无霸汉堡和 6 个小皇堡需要 4*1 + 2*6 = 16 片番茄和 1 + 6 = 7 片奶酪。不会剩下原料。

示例 2：

输入：tomatoSlices = 17, cheeseSlices = 4
输出：[]
解释：只制作小皇堡和巨无霸汉堡无法用光全部原料。

示例 3：

输入：tomatoSlices = 4, cheeseSlices = 17
输出：[]
解释：制作 1 个巨无霸汉堡会剩下 16 片奶酪，制作 2 个小皇堡会剩下 15 片奶酪。

示例 4：

输入：tomatoSlices = 0, cheeseSlices = 0
输出：[0,0]

示例 5：

输入：tomatoSlices = 2, cheeseSlices = 1
输出：[0,1]

提示：

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

lightbulb

解题思路

方法一：数学

我们设巨无霸汉堡数量为 $x$ ，小皇堡数量为 $y$ ，则有：

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

将上述两式转换，可以得到：

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

其中 $x$ 和 $y$ 必须为非负整数。

时间复杂度 $O(1)$ ，空间复杂度 $O(1)$ 。

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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]

speed

复杂度分析

指标	值
时间	complexity is O(1) because calculations involve direct arithmetic without iteration. Space complexity is O(1) since only a few integer variables are stored, regardless of input size.
空间	Depends on the final approach

psychology

面试官常问的追问

外企场景

question_mark
Look for direct algebraic formulation rather than iterative brute force.
question_mark
Check candidate solutions against ingredient parity and zero-remainder constraints.
question_mark
Consider edge cases where tomatoSlices or cheeseSlices are zero or insufficient.

warning

常见陷阱

外企场景

error
Failing to check that tomatoSlices must be even for a valid solution.
error
Assuming negative counts are valid for jumbo or small burgers.
error
Ignoring integer division issues leading to fractional burger counts.

swap_horiz

进阶变体

外企场景

arrow_right_alt
Varying burger ingredient ratios, e.g., jumbo uses 5 tomato slices instead of 4.
arrow_right_alt
Limiting total number of burgers available, requiring optimization within constraints.
arrow_right_alt
Adding multiple burger types beyond jumbo and small with unique ingredient requirements.

help

常见问题

外企场景

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