LeetCode 1276 - Number of Burgers with No Waste of Ingredients

Here is a complete, detailed technical solution guide for LeetCode 1276 - Number of Burgers with No Waste of Ingredients

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LeetCode Problem 1276

Difficulty: 🟡 Medium
Topics: Math

Solution

Here is a complete, detailed technical solution guide for LeetCode 1276 - Number of Burgers with No Waste of Ingredients, following your requested format.

Problem Understanding

The problem asks us to determine how many Jumbo and Small burgers we can make using exactly a given number of tomato slices and cheese slices without leaving any leftover ingredients. Each Jumbo Burger requires 4 tomato slices and 1 cheese slice, while each Small Burger requires 2 tomato slices and 1 cheese slice. We are given two integers, tomatoSlices and cheeseSlices, representing the total available ingredients. Our task is to return [total_jumbo, total_small] that consumes all the ingredients. If no combination exists that uses up all ingredients exactly, we return an empty list [].

Restated in algebraic terms, let J be the number of Jumbo burgers and S be the number of Small burgers. Then the problem reduces to solving the system:

4 * J + 2 * S = tomatoSlices
1 * J + 1 * S = cheeseSlices

The solution must satisfy J >= 0 and S >= 0, and both numbers must be integers.

Constraints tell us 0 <= tomatoSlices, cheeseSlices <= 10^7, which allows for O(1) arithmetic solutions but makes brute force iteration infeasible for large inputs.

Important edge cases include when either tomatoSlices or cheeseSlices is zero, when ingredients cannot match any combination (e.g., odd tomato counts), and when one type of burger dominates.

Approaches

Brute Force Approach:

A naive approach would be to iterate over all possible numbers of Jumbo burgers J from 0 to tomatoSlices // 4 and, for each, check if the remaining tomato slices and cheese slices can form an integer number of Small burgers S. While this guarantees correctness, it is inefficient because in the worst case it may require up to 2.5 million iterations (for tomatoSlices = 10^7), which is unnecessary since this problem can be solved with simple algebra.

Optimal Approach:

The key insight is to treat this as a system of linear equations. From the cheese equation J + S = cheeseSlices, we can express S = cheeseSlices - J. Substituting into the tomato equation:

4 * J + 2 * (cheeseSlices - J) = tomatoSlices
4J + 2*cheeseSlices - 2J = tomatoSlices
2J + 2*cheeseSlices = tomatoSlices
2J = tomatoSlices - 2*cheeseSlices
J = (tomatoSlices - 2*cheeseSlices) / 2

Once J is computed, S = cheeseSlices - J. The solution is valid only if J >= 0, S >= 0, and both are integers. If these conditions are not satisfied, no solution exists.

Approach Time Complexity Space Complexity Notes
Brute Force O(n) O(1) Iterate all possible Jumbo burgers and check Small burgers
Optimal O(1) O(1) Solve system of linear equations algebraically

Algorithm Walkthrough

  1. Check feasibility: First, ensure that tomatoSlices is at least twice the cheeseSlices (otherwise we can't satisfy the minimal 2-tomato requirement for each cheese) and is even. If tomatoSlices is odd, no integer solution exists.
  2. Compute Jumbo burgers: Use the derived formula J = (tomatoSlices - 2 * cheeseSlices) // 2.
  3. Compute Small burgers: Calculate S = cheeseSlices - J.
  4. Validate solution: Check that J >= 0 and S >= 0. If either is negative, return [].
  5. Return result: Return [J, S].

Why it works:

The algorithm works because it directly solves the linear system that describes the problem. There is only one valid integer solution for the system if it exists. The algebraic derivation ensures that we check the necessary and sufficient conditions for no ingredient waste.

Python Solution

from typing import List

class Solution:
    def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:
        if tomatoSlices % 2 != 0 or tomatoSlices < 2 * cheeseSlices or tomatoSlices > 4 * cheeseSlices:
            return []
        
        jumbo = (tomatoSlices - 2 * cheeseSlices) // 2
        small = cheeseSlices - jumbo
        
        if jumbo < 0 or small < 0:
            return []
        
        return [jumbo, small]

Explanation:

The solution first checks impossible cases: odd tomatoSlices cannot be split evenly between 2 and 4 per burger, and tomatoSlices outside [2 * cheeseSlices, 4 * cheeseSlices] cannot form a valid combination. Then, it calculates jumbo using the derived formula and derives small accordingly. A final check ensures non-negativity.

Go Solution

func numOfBurgers(tomatoSlices int, cheeseSlices int) []int {
    if tomatoSlices%2 != 0 || tomatoSlices < 2*cheeseSlices || tomatoSlices > 4*cheeseSlices {
        return []int{}
    }

    jumbo := (tomatoSlices - 2*cheeseSlices) / 2
    small := cheeseSlices - jumbo

    if jumbo < 0 || small < 0 {
        return []int{}
    }

    return []int{jumbo, small}
}

Go-specific Notes:

Go uses slices to return results. The check for odd numbers uses modulo %. Since Go does not throw exceptions for out-of-bound slices like Python, we safely return an empty slice when the solution is invalid. Integer division automatically floors the result, matching Python's // behavior.

Worked Examples

Example 1: tomatoSlices = 16, cheeseSlices = 7

  1. Check feasibility: 16 is even, 16 ≥ 2_7=14, 16 ≤ 4_7=28 → OK
  2. Compute Jumbo: J = (16 - 2*7) // 2 = (16 - 14) // 2 = 1
  3. Compute Small: S = 7 - 1 = 6
  4. Validate: J=1 ≥ 0, S=6 ≥ 0 → OK
  5. Return [1, 6]

Example 2: tomatoSlices = 17, cheeseSlices = 4

  1. 17 is odd → immediate return []

Example 3: tomatoSlices = 4, cheeseSlices = 17

  1. Check feasibility: 4 < 2*17=34 → impossible → return []

Complexity Analysis

Measure Complexity Explanation
Time O(1) Only arithmetic operations and constant number of checks
Space O(1) Only a few variables used for computation, output size is constant

The optimal solution does not iterate over any range; it computes results directly, making it extremely efficient.

Test Cases

# Provided examples
assert Solution().numOfBurgers(16, 7) == [1, 6]  # exact solution exists
assert Solution().numOfBurgers(17, 4) == []     # odd tomatoSlices
assert Solution().numOfBurgers(4, 17) == []     # insufficient tomatoSlices

# Boundary and edge cases
assert Solution().numOfBurgers(0, 0) == [0, 0]  # no ingredients
assert Solution().numOfBurgers(2, 1) == [0, 1]  # smallest small burger
assert Solution().numOfBurgers(4, 1) == [1, 0]  # smallest jumbo burger
assert Solution().numOfBurgers(6, 1) == [0, 1]  # impossible: 6 tomato, 1 cheese cannot split evenly
assert Solution().numOfBurgers(10000000, 2500000) == [0, 2500000]  # large inputs, small-only solution
assert Solution().numOfBurgers(10000000, 3000000) == [1000000, 2000000]  # large inputs, mixed solution
Test Why
[16, 7] Standard case with mixed burgers
[17, 4] Odd tomatoSlices, impossible
[4, 17] Not enough tomatoSlices
[0, 0] No ingredients, edge case
[2, 1] Minimum small burger
[4, 1] Minimum jumbo burger
[6, 1] Invalid combination
[10^7, 2.5*10^6] Large input, small-only solution
[10^7, 3*10^6] Large input, mixed solution

Edge Cases

  1. Odd tomato slices: Since both burger types require even numbers of tomato slices