How do you solve Number of Burgers with No Waste of Ingredients on LeetCode?

Number of Burgers with No Waste of Ingredients (LeetCode #1276) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: If tomatoSlices is odd, it's impossible to create burgers without waste since both burger types require an even number of tomato slices.

What is the time complexity of Number of Burgers with No Waste of Ingredients?

The optimal solution for Number of Burgers with No Waste of Ingredients runs in O(1) time and uses O(1) space. The time complexity is O(1) because we are performing a constant number of arithmetic operations regardless of the input size.

What is the brute force solution for Number of Burgers with No Waste of Ingredients?

The brute force approach for Number of Burgers with No Waste of Ingredients is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves trying every possible combination of jumbo and small burgers to see if we can use up all the tomato and cheese slices. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Number of Burgers with No Waste of Ingredients?

Number of Burgers with No Waste of Ingredients uses the following concepts: Math. The recommended patterns to study are: Mathematical derivation for optimization, Linear equations and inequalities.

What are the interview tips for Number of Burgers with No Waste of Ingredients?

Always check edge cases, such as when tomatoSlices or cheeseSlices is zero. Be clear about your assumptions and constraints before diving into coding. Explain your thought process as you work through the problem; this shows your understanding and can help you catch mistakes.

What are common mistakes when solving Number of Burgers with No Waste of Ingredients?

Failing to check if the calculated number of burgers is non-negative. Not considering the parity of tomatoSlices when determining if a solution is possible.

#1276

Number of Burgers with No Waste of Ingredients

Medium

MathMathematical derivation for optimizationLinear equations and inequalities

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

💡

Intuition

Time O(1)Space O(1)

Instead of brute-forcing through all combinations, we can derive a mathematical relationship between the number of jumbo and small burgers based on the total slices. This allows us to directly compute the counts without iteration.

⚙️

Algorithm

4 steps

1Step 1: Use the equations: 4 * jumbo + 2 * small = tomatoSlices and jumbo + small = cheeseSlices.
2Step 2: Rearrange these equations to express small in terms of jumbo: small = cheeseSlices - jumbo.
3Step 3: Substitute this into the first equation and solve for jumbo: jumbo = (tomatoSlices - 2 * cheeseSlices) / 2.
4Step 4: Calculate small using the derived jumbo value and validate both counts are non-negative.

solution.py6 lines

1def numOfBurgers(tomatoSlices, cheeseSlices):
2    jumbo = (tomatoSlices - 2 * cheeseSlices) // 2
3    small = cheeseSlices - jumbo
4    if jumbo < 0 or small < 0 or (tomatoSlices - 2 * cheeseSlices) % 2 != 0:
5        return []
6    return [jumbo, small]

ℹ

Complexity note: The time complexity is O(1) because we are performing a constant number of arithmetic operations regardless of the input size.

1If tomatoSlices is odd, it's impossible to create burgers without waste since both burger types require an even number of tomato slices.
2The number of cheese slices must be sufficient to create the burgers derived from the tomato slices.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.