Number of Burgers with No Waste of Ingredients solution and explanation
The problem
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.
def numOfBurgers(tomatoSlices, cheeseSlices):
if tomatoSlices % 2 != 0 or cheeseSlices * 2 > tomatoSlices or cheeseSlices * 4 < tomatoSlices:
return []
jumbo = tomatoSlices // 2 - cheeseSlices
small = cheeseSlices - jumbo
return [jumbo, small]
print(numOfBurgers(16, 7))import java.util.*;
public class Burgers {
public List<Integer> numOfBurgers(int tomatoSlices, int cheeseSlices) {
if (tomatoSlices % 2 != 0 || cheeseSlices * 2 > tomatoSlices || cheeseSlices * 4 < tomatoSlices) {
return new ArrayList<>();
}
int jumbo = (tomatoSlices / 2) - cheeseSlices;
int small = cheeseSlices - jumbo;
return Arrays.asList(jumbo, small);
}
public static void main(String[] args) {
Burgers b = new Burgers();
System.out.println(b.numOfBurgers(16, 7));
}
}#include <iostream>
#include <vector>
using namespace std;
vector<int> numOfBurgers(int tomatoSlices, int cheeseSlices) {
if (tomatoSlices % 2 != 0 || cheeseSlices * 2 > tomatoSlices || cheeseSlices * 4 < tomatoSlices) {
return {};
}
int jumbo = (tomatoSlices / 2) - cheeseSlices;
int small = cheeseSlices - jumbo;
return {jumbo, small};
}
int main() {
for (auto& result : numOfBurgers(16, 7)) {
cout << result << " ";
}
return 0;
}function numOfBurgers(tomatoSlices, cheeseSlices) {
if (tomatoSlices % 2 !== 0 || cheeseSlices * 2 > tomatoSlices || cheeseSlices * 4 < tomatoSlices) {
return [];
}
let jumbo = tomatoSlices / 2 - cheeseSlices;
let small = cheeseSlices - jumbo;
return [jumbo, small];
}
console.log(numOfBurgers(16, 7));#include <stdio.h>
void burgers(int tomatoSlices, int cheeseSlices) {
if (tomatoSlices % 2 != 0 || cheeseSlices * 2 > tomatoSlices || cheeseSlices * 4 < tomatoSlices) {
printf("[]\n");
return;
}
int jumbo = (tomatoSlices / 2) - cheeseSlices;
int small = cheeseSlices - jumbo;
printf("[%d, %d]\n", jumbo, small);
}
int main() {
burgers(16, 7);
burgers(17, 4);
burgers(4, 17);
return 0;
}using System;
using System.Collections.Generic;
public class Program {
public static List<int> NumOfBurgers(int tomatoSlices, int cheeseSlices) {
if (tomatoSlices % 2 != 0 || cheeseSlices * 2 > tomatoSlices || cheeseSlices * 4 < tomatoSlices) {
return new List<int>();
}
int jumbo = tomatoSlices / 2 - cheeseSlices;
int small = cheeseSlices - jumbo;
return new List<int> { jumbo, small };
}
public static void Main() {
var result = NumOfBurgers(16, 7);
result.ForEach(Console.WriteLine);
}
}This Python function uses integer arithmetic to find solutions for the number of jumbo and small burgers, if a solution can be constructed, based on given tomato and cheese slice constraints.
Time Complexity: O(1), simple calculations are employed.
Space Complexity: O(1), minimal space required to store results.
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.
def iterativeSolution(tomatoSlices, cheeseSlices):
for jumbo in range(cheeseSlices + 1):
small = cheeseSlices - jumbo
if 4 * jumbo + 2 * small == tomatoSlices:
return [jumbo, small]
return []
print(iterativeSolution(16, 7))import java.util.*;
public class BurgersIterative {
public List<Integer> iterativeSolution(int tomatoSlices, int cheeseSlices) {
for (int jumbo = 0; jumbo <= cheeseSlices; ++jumbo) {
int small = cheeseSlices - jumbo;
if (4 * jumbo + 2 * small == tomatoSlices) {
return Arrays.asList(jumbo, small);
}
}
return new ArrayList<>();
}
public static void main(String[] args) {
BurgersIterative bi = new BurgersIterative();
System.out.println(bi.iterativeSolution(16, 7));
}
}#include <iostream>
#include <vector>
using namespace std;
vector<int> iterativeSolution(int tomatoSlices, int cheeseSlices) {
for (int jumbo = 0; jumbo <= cheeseSlices; ++jumbo) {
int small = cheeseSlices - jumbo;
if (4 * jumbo + 2 * small == tomatoSlices) {
return {jumbo, small};
}
}
return {};
}
int main() {
vector<int> result = iterativeSolution(16, 7);
for (auto& res : result) {
cout << res << " ";
}
return 0;
}function iterativeSolution(tomatoSlices, cheeseSlices) {
for (let jumbo = 0; jumbo <= cheeseSlices; jumbo++) {
let small = cheeseSlices - jumbo;
if (4 * jumbo + 2 * small === tomatoSlices) {
return [jumbo, small];
}
}
return [];
}
console.log(iterativeSolution(16, 7));#include <stdio.h>
void iterativeSolution(int tomatoSlices, int cheeseSlices) {
for (int jumbo = 0; jumbo <= cheeseSlices; ++jumbo) {
int small = cheeseSlices - jumbo;
if (4 * jumbo + 2 * small == tomatoSlices) {
printf("[%d, %d]\n", jumbo, small);
return;
}
}
printf("[]\n");
}
int main() {
iterativeSolution(16, 7);
return 0;
}using System;
using System.Collections.Generic;
public class Program {
public static List<int> IterativeSolution(int tomatoSlices, int cheeseSlices) {
for (int jumbo = 0; jumbo <= cheeseSlices; ++jumbo) {
int small = cheeseSlices - jumbo;
if (4 * jumbo + 2 * small == tomatoSlices) {
return new List<int> { jumbo, small };
}
}
return new List<int>();
}
public static void Main() {
var result = IterativeSolution(16, 7);
result.ForEach(Console.WriteLine);
}
}The iterative function loops over jumbo burger possibilities and checks their matching with small values under specified conditions.
Time Complexity: O(n) due to the single linear iteration up to the number of cheeseSlices.
Space Complexity: O(1) as no additional data structures are stored aside from the loop variable.
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)$.
function numOfBurgers(tomatoSlices: number, cheeseSlices: number): number[] {
const k = 4 * cheeseSlices - tomatoSlices;
const y = k >> 1;
const x = cheeseSlices - y;
return k % 2 || y < 0 || x < 0 ? [] : [x, y];
}func numOfBurgers(tomatoSlices int, cheeseSlices int) []int {
k := 4*cheeseSlices - tomatoSlices
y := k / 2
x := cheeseSlices - y
if k%2 != 0 || x < 0 || y < 0 {
return []int{}
}
return []int{x, y}
}impl Solution {
pub fn num_of_burgers(tomato_slices: i32, cheese_slices: i32) -> Vec<i32> {
let k = 4 * cheese_slices - tomato_slices;
let y = k / 2;
let x = cheese_slices - y;
if k % 2 != 0 || y < 0 || x < 0 {
Vec::new()
} else {
vec![x, y]
}
}
}Complexity comparison
| Approach | Complexity |
|---|---|
| Directly Solve Using Equations | Time Complexity: O(1), simple calculations are employed. Space Complexity: O(1), minimal space required to store results. |
| Iterative Check for Solution | Time Complexity: O(n) due to the single linear iteration up to the number of Space Complexity: O(1) as no additional data structures are stored aside from the loop variable. |
| Mathematics | – |
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 |
Frequently asked questions
Is Number of Burgers with No Waste of Ingredients easy or hard?
The problem is rated Medium on LeetCode. Implementation is simple, but recognizing that the constraints form a solvable system of equations requires some mathematical insight.
Number of Burgers with No Waste of Ingredients Python/Java solution
Most implementations compute jumbo burgers using the derived formula and then calculate small burgers from the remaining cheese slices. The logic is identical across Python, Java, C++, C#, and JavaScript because it only involves integer arithmetic and conditional checks.
How to solve Number of Burgers with No Waste of Ingredients in O(1)?
Model the burger counts with equations. Let j be jumbo burgers and s be small burgers. From the constraints: 4j + 2s = tomatoSlices and j + s = cheeseSlices. Solve the system to compute j and s directly, then verify both are non-negative integers that satisfy the equations.
What is the best approach for Number of Burgers with No Waste of Ingredients?
The best approach is solving the ingredient constraints as a system of equations. Using 4j + 2s = tomatoSlices and j + s = cheeseSlices, you can compute the number of jumbo and small burgers directly in O(1) time and O(1) space. This avoids iteration and guarantees optimal performance.
Is Number of Burgers with No Waste of Ingredients asked at Google/Amazon/Meta?
This type of problem appears in interviews at companies like Amazon, Google, and Meta when testing mathematical reasoning and constraint modeling. It is categorized as a Medium problem because the key challenge is recognizing the equation relationship rather than implementing complex data structures.
What data structure is used in Number of Burgers with No Waste of Ingredients?
No special data structure is required. The problem is purely mathematical and relies on solving linear equations and validating integer constraints. Only a few integer variables are used to compute the result.
What is the time complexity of Number of Burgers with No Waste of Ingredients?
The optimal equation-based solution runs in O(1) time because it performs a fixed number of arithmetic operations. Space complexity is also O(1). A brute-force or iterative approach that checks possible jumbo burger counts takes O(n) time where n is roughly tomatoSlices / 4.