Number of Burgers with No Waste of Ingredients

Number of Burgers with No Waste of Ingredients - Problem

Math Medium

You're managing a burger restaurant and need to determine the exact combination of burgers to make with your available ingredients to avoid any waste.

Your restaurant serves two types of burgers:

Jumbo Burger: Requires 4 tomato slices and 1 cheese slice
Small Burger: Requires 2 tomato slices and 1 cheese slice

Given tomatoSlices and cheeseSlices available in your kitchen, you need to find how many of each burger type to make such that:

All tomato slices are used (no waste)
All cheese slices are used (no waste)

Goal: Return [total_jumbo, total_small] representing the number of Jumbo and Small burgers respectively. If it's impossible to use all ingredients exactly, return an empty array [].

Example: With 16 tomato slices and 7 cheese slices, you can make 1 Jumbo burger (uses 4 tomatoes, 1 cheese) and 6 Small burgers (uses 12 tomatoes, 6 cheese), using all ingredients perfectly.

Input & Output

example_1.py — Basic Valid Case

$ Input: tomatoSlices = 16, cheeseSlices = 7

› Output: [1, 6]

💡 Note: 1 Jumbo burger uses 4 tomatoes + 1 cheese, 6 Small burgers use 12 tomatoes + 6 cheese. Total: 16 tomatoes, 7 cheese - perfect match!

example_2.py — Impossible Case

$ Input: tomatoSlices = 17, cheeseSlices = 4

› Output: []

💡 Note: No valid combination exists. Even with 4 Jumbo burgers (16 tomatoes, 4 cheese), we have 1 tomato left over with no cheese to make another burger.

example_3.py — All Small Burgers

$ Input: tomatoSlices = 4, cheeseSlices = 2

› Output: [0, 2]

💡 Note: Make 2 Small burgers using exactly 4 tomato slices and 2 cheese slices. No Jumbo burgers needed.

Constraints

0 ≤ tomatoSlices, cheeseSlices ≤ 10⁷
Time limit: 1 second per test case
Both inputs are non-negative integers

Visualization

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Understanding the Visualization

Identify the constraints

Each cheese slice makes exactly one burger, tomatoes vary by burger type

Set up the math

Create equations representing ingredient usage for each burger type

Solve algebraically

Use substitution to solve for the exact number of each burger type

Validate the solution

Ensure both burger counts are non-negative integers

Key Takeaway

🎯 Key Insight: This is a disguised system of linear equations problem - instead of trying all combinations, we can solve mathematically in O(1) time using algebra!

Asked in

G Google 15 ⊞ Microsoft 12 a Amazon 8 f Meta 6

The optimal approach treats this as a system of linear equations: 4J + 2S = T and J + S = C. Solving algebraically gives us J = (T - 2C) / 2 and S = C - J in O(1) time, eliminating the need for any loops or iterations.

Common Approaches

Approach	Time	Space	Notes
✓ Mathematical Solution (Optimal)	O(1)	O(1)	Solve the system of linear equations directly using algebra

Mathematical Solution (Optimal) — Algorithm Steps

Set up equations: 4J + 2S = T and J + S = C
Solve for J: J = (T - 2C) / 2
Calculate S: S = C - J
Validate: check if J ≥ 0, S ≥ 0, and both are integers

Visualization

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Step-by-Step Walkthrough

Set up equations

4J + 2S = T and J + S = C

Substitute and solve

Replace S with (C-J) to get J = (T-2C)/2

Calculate values

Find J first, then S = C - J

Validate solution

Check if both values are non-negative integers

Code -

solution.c — C

#include <stdlib.h>

int* numOfBurgers(int tomatoSlices, int cheeseSlices, int* returnSize) {
    *returnSize = 0;
    
    // Mathematical solution: solve system of equations
    // 4J + 2S = T (tomatoes)
    // J + S = C (cheese)
    // Solving: J = (T - 2C) / 2, S = C - J
    
    // Check if (T - 2C) is even for integer solution
    if ((tomatoSlices - 2 * cheeseSlices) % 2 != 0) {
        return NULL;
    }
    
    int jumbo = (tomatoSlices - 2 * cheeseSlices) / 2;
    int small = cheeseSlices - jumbo;
    
    // Validate non-negative solution
    if (jumbo >= 0 && small >= 0) {
        int* result = (int*)malloc(2 * sizeof(int));
        result[0] = jumbo;
        result[1] = small;
        *returnSize = 2;
        return result;
    }
    
    return NULL;
}

Time & Space Complexity

Time Complexity

⏱️

O(1)

Direct mathematical calculation with constant time operations

✓ Linear Growth

Space Complexity

O(1)

Only using constant space for calculation variables

✓ Linear Space

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Mathematical Solution (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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