# Solving Leetcode Interviews in Seconds with AI: Number of Burgers with No Waste of Ingredients


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1276" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:  Jumbo Burger: 4 tomato slices and 1 cheese slice. Small Burger: 2 Tomato slices and 1 cheese 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  

	# Explanation
	Here's a solution to the burger problem:

*   **Approach:** We can formulate two equations based on the given information. Let `j` be the number of jumbo burgers and `s` be the number of small burgers. Then, we have: 4j + 2s = tomatoSlices and j + s = cheeseSlices. Solving these equations for `j` and `s` will give us the answer. We can solve for `j` in the second equation (j = cheeseSlices - s) and substitute it into the first equation. This gives us 4(cheeseSlices - s) + 2s = tomatoSlices, which simplifies to 4 * cheeseSlices - 2s = tomatoSlices. Thus, 2s = 4 * cheeseSlices - tomatoSlices, and s = (4 * cheeseSlices - tomatoSlices) / 2. We then solve for j = cheeseSlices - s. We need to check if the calculated values of `j` and `s` are non-negative integers to ensure a valid solution.

*   **Complexity:** The runtime complexity is O(1) as the solution involves only arithmetic operations. The storage complexity is also O(1) as we use a fixed number of variables.

	
	# Code
	```python
	def solve():
    tomatoSlices = int(input())
    cheeseSlices = int(input())
    jumbo = (tomatoSlices - 2 * cheeseSlices) / 2
    small = cheeseSlices - jumbo
    if jumbo >= 0 and small >= 0 and jumbo.is_integer() and small.is_integer():
        print([int(jumbo), int(small)])
    else:
        print([])

solve()
	```