Number of Burgers with No Waste of Ingredients - Solution & Explanation
Problem Statement
Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:
- Jumbo Burger:
4tomato slices and1cheese slice. - Small Burger:
2Tomato slices and1cheese 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
Approach Overview
Problem Overview: You are given the number of tomatoSlices and cheeseSlices. A jumbo burger uses 4 tomato slices and 1 cheese slice, while a small burger uses 2 tomato slices and 1 cheese slice. Your task is to determine how many jumbo and small burgers can be made so that all ingredients are used with no leftovers.
The challenge reduces to solving a small system of equations. Each burger consumes exactly one cheese slice, but tomato usage differs between burger types. If a valid combination exists, return the number of jumbo and small burgers; otherwise return an empty result.
Approach 1: Iterative Check for Solution (O(n) time, O(1) space)
The straightforward approach is to try all possible counts of jumbo burgers. Since each jumbo burger uses 4 tomato slices, the maximum possible jumbo burgers is tomatoSlices / 4. For every candidate count, compute the remaining tomato and cheese slices and check whether they can form small burgers.
The validation step is simple: remaining tomatoes must equal 2 * smallBurgers, and remaining cheese slices must match the total burgers used. Iterate through feasible jumbo counts and return the first valid pair. This method relies on basic arithmetic and loops, making it easy to reason about but slightly inefficient when the tomato count is large.
This approach mainly exercises careful iteration and arithmetic validation. It connects well with problems under math and basic constraint checking.
Approach 2: Directly Solve Using Equations (O(1) time, O(1) space)
The optimal solution models the burger counts as two variables. Let j be jumbo burgers and s be small burgers. From the ingredient rules you get two equations:
4j + 2s = tomatoSlicesj + s = cheeseSlices
Solving this system eliminates one variable. Substitute s = cheeseSlices - j into the tomato equation and simplify to compute j. Once j is known, calculate s. The only remaining checks ensure both values are non‑negative integers and satisfy the original constraints.
This approach turns the problem into a small algebra exercise and avoids iteration entirely. The runtime becomes constant because only a few arithmetic operations are performed regardless of input size. Problems like this commonly appear in math and equation solving categories where recognizing relationships between variables leads directly to the optimal solution.
Recommended for interviews: The equation-based approach is what interviewers expect. It demonstrates that you can translate constraints into algebra and derive a constant-time solution. The iterative approach still shows good reasoning and can serve as a stepping stone, but the O(1) equation solution highlights stronger problem‑solving skill.
Approach 1: Directly Solve Using Equations
To solve the problem, we can formulate it into a set of linear equations. Let jumboCount be the number of jumbo burgers and smallCount be the number of small burgers. The two equations based on problem statement are:
- 4 *
jumboCount+ 2 *smallCount=tomatoSlices jumboCount+smallCount=cheeseSlices
By solving these simultaneously, we can derive the values for jumboCount and smallCount.
For valid solutions, jumboCount and smallCount need to be non-negative integers. If a solution exists, return it in the form of a list containing [jumboCount, smallCount]. Otherwise, return an empty list.
This C function calculates the number of jumbo and small burgers by solving the linear equations derived from the number of given ingredients.
The code checks if it's possible to solve this using the tomatoSlices divisibility, and bounds checks with cheeseSlices. If it isn't possible, an empty array is returned.
Complexity
Time Complexity: O(1) because it's a simple mathematical computation.
Space Complexity: O(1) since no extra space is used.
Approach 2: Iterative Check for Solution
Iterative Approach: This method involves iterating to find solutions. You keep trying different values of jumbo until consistency with constants and checks for valid small counts are verified.
Drawback: This is typically less efficient as it explicitly loops through possible combinations but is a straightforward means for those starting out in problem-solving processes.
The code iterates through potential values for the jumbo burger count while validating possibilities for the total given conditions.
This method serves as an alternate but less optimal way to determine combinations valid for input constraints.
Complexity
Time Complexity: O(n), where n is limited by cheeseSlices.
Space Complexity: O(1), only uses local variables.
Approach 3: 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)$.
Complexity Comparison
| Approach | Complexity |
|---|---|
| Directly Solve Using Equations | Time Complexity: O(1) because it's a simple mathematical computation. Space Complexity: O(1) since no extra space is used. |
| Iterative Check for Solution | Time Complexity: O(n), where n is limited by Space Complexity: O(1), only uses local variables. |
| Mathematics | — |
Detailed Complexity Analysis
| Approach | Time | Space | When to Use |
|---|---|---|---|
| Iterative Check for Solution | O(n) | O(1) | Useful for understanding constraints or when deriving the formula is not obvious |
| Directly Solve Using Equations | O(1) | O(1) | Best approach for interviews and production due to constant time and simple arithmetic |
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