LeetCode in Kotlin
1276. Number of Burgers with No Waste of Ingredients
Medium
Given two integers tomatoSlices and cheeseSlices. The ingredients of different burgers are as follows:
- Jumbo Burger:
4tomato slices and1cheese slice. - Small Burger:
2Tomato slices and1cheese 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
Solution
class Solution {
fun numOfBurgers(tomatoSlices: Int, cheeseSlices: Int): List<Int> {
val numbers: MutableList<Int> = ArrayList()
val numberOfCheese = cheeseSlices * 4
val remaining = numberOfCheese - tomatoSlices
if (remaining >= 0 && remaining % 2 != 1) {
val numberOfSmall = remaining / 2
val numberOfLarge = cheeseSlices - numberOfSmall
if (numberOfLarge < 0) {
return numbers
}
numbers.add(numberOfLarge)
numbers.add(numberOfSmall)
}
return numbers
}
}