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:
4
tomato slices and1
cheese slice.Small Burger:
2
Tomato slices and1
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
class Solution:
def numOfBurgers(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:
# Got this formula using equations 4*x+2*y = tomatoSlices , x+y=cheeseSlices
y = (2*cheeseSlices - tomatoSlices/2)
x = (cheeseSlices-y)
if x < 0 or y < 0 or x%1 or y%1:
return []
return [int(x),int(y)]
# TLE with test case 1000000 1000000
def numOfBurgersOld(self, tomatoSlices: int, cheeseSlices: int) -> List[int]:
@lru_cache()
def recursion(tomatoSlices, cheeseSlices, jumbo, small):
if tomatoSlices == 0 and cheeseSlices == 0:
return [jumbo, small]
elif tomatoSlices < 0 or cheeseSlices < 0:
# print(jumbo, small)
return []
r1 = recursion(tomatoSlices-4, cheeseSlices-1, jumbo + 1, small)
if r1 :
return r1
r2 = recursion(tomatoSlices-2, cheeseSlices-1, jumbo, small+1)
return r2
result = recursion(tomatoSlices, cheeseSlices, 0, 0)
return result
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