forked from shuboc/LeetCode-2
-
Notifications
You must be signed in to change notification settings - Fork 0
/
shopping-offers.py
60 lines (57 loc) · 2.63 KB
/
shopping-offers.py
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
# Time: O(n * 2^n)
# Space: O(n)
# In LeetCode Store, there are some kinds of items to sell. Each item has a price.
#
# However, there are some special offers, and a special offer consists of one or
# more different kinds of items with a sale price.
#
# You are given the each item's price, a set of special offers,
# and the number we need to buy for each item. The job is to output the lowest price you have to pay
# for exactly certain items as given, where you could make optimal use of the special offers.
#
# Each special offer is represented in the form of an array,
# the last number represents the price you need to pay for this special offer,
# other numbers represents how many specific items you could get if you buy this offer.
#
# You could use any of special offers as many times as you want.
#
# Example 1:
# Input: [2,5], [[3,0,5],[1,2,10]], [3,2]
# Output: 14
# Explanation:
# There are two kinds of items, A and B. Their prices are $2 and $5 respectively.
# In special offer 1, you can pay $5 for 3A and 0B
# In special offer 2, you can pay $10 for 1A and 2B.
# You need to buy 3A and 2B, so you may pay $10 for 1A and 2B (special offer #2), and $4 for 2A.
# Example 2:
# Input: [2,3,4], [[1,1,0,4],[2,2,1,9]], [1,2,1]
# Output: 11
# Explanation:
# The price of A is $2, and $3 for B, $4 for C.
# You may pay $4 for 1A and 1B, and $9 for 2A ,2B and 1C.
# You need to buy 1A ,2B and 1C, so you may pay $4 for 1A and 1B (special offer #1), and $3 for 1B, $4 for 1C.
# You cannot add more items, though only $9 for 2A ,2B and 1C.
# Note:
# There are at most 6 kinds of items, 100 special offers.
# For each item, you need to buy at most 6 of them.
# You are not allowed to buy more items than you want, even if that would lower the overall price.
class Solution(object):
def shoppingOffers(self, price, special, needs):
"""
:type price: List[int]
:type special: List[List[int]]
:type needs: List[int]
:rtype: int
"""
def shoppingOffersHelper(price, special, needs, i):
if i == len(special):
return sum(map(lambda x, y: x*y, price, needs))
result = shoppingOffersHelper(price, special, needs, i+1)
for j in xrange(len(needs)):
needs[j] -= special[i][j]
if all(need >= 0 for need in needs):
result = min(result, special[i][-1] + shoppingOffersHelper(price, special, needs, i))
for j in xrange(len(needs)):
needs[j] += special[i][j]
return result
return shoppingOffersHelper(price, special, needs, 0)