README.md

1276. Number of Burgers with No Waste of Ingredients

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] Explanation: 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: [] Explanation: There will be no way to use all ingredients to make small and jumbo burgers. 

Example 3:

Input: tomatoSlices = 4, cheeseSlices = 17 Output: [] Explanation: Making 1 jumbo burger there will be 16 cheese remaining and making 2 small burgers there will be 15 cheese remaining. 

Example 4:

Input: tomatoSlices = 0, cheeseSlices = 0 Output: [0,0] 

Example 5:

Input: tomatoSlices = 2, cheeseSlices = 1 Output: [0,1] 

Constraints:

• 0 <= tomatoSlices <= 10^7
• 0 <= cheeseSlices <= 10^7

Solutions (Python)

1. Mathematical

classSolution: defnumOfBurgers(self, tomatoSlices: int, cheeseSlices: int) ->List[int]: iftomatoSlices%2==0 \ andtomatoSlices>=2*cheeseSlices \ andtomatoSlices<=4*cheeseSlices: return [ tomatoSlices//2-cheeseSlices, 2*cheeseSlices-tomatoSlices//2 ] return []

Solutions (Rust)

1. Mathematical

implSolution{pubfnnum_of_burgers(tomato_slices:i32,cheese_slices:i32) -> Vec<i32>{if tomato_slices % 2 == 0 && tomato_slices >= 2* cheese_slices && tomato_slices <= 4* cheese_slices {vec![ tomato_slices / 2 - cheese_slices,2* cheese_slices - tomato_slices / 2,]}else{vec![]}}}