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+# 638. Shopping Offers
+
+"""
+in leetcode store, there are 'n' 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 an integer
+array 'price' where 'price[i]' is the price of the i'th item, and an integer
+array 'needs' where 'needs[i]' is the number of pieces the i'th item you want
+to buy. you are also given an array 'special', where 'special[i]' is of size
+'n + 1' where 'special[i][j]' is the number of pieces of the j'th item in the
+i'th other. return the lowest price you have to pay for exactly certain items
+as given, where you could make optimal use of the special offers. you are not
+allowed to buy more items that 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
+ """
+ dp = {}
+
+ def dfs(curr):
+ val = sum(curr[i] * price[i] for i in range(len(needs)))
+ for *spec, cost in special:
+ tmp = [curr[j] - spec[j] for j in range(len(needs))]
+ if min(tmp) >= 0: # skip deals that exceed needs
+ val = min(val, dp.get(tuple(tmp), dfs(tmp)) + cost)
+ dp[tuple(curr)] = val
+ return val
+
+ return dfs(needs)
+
+
+if __name__ == "__main__":
+ obj = Solution()
+ print(
+ obj.shoppingOffers(price=[2, 5], special=[[3, 0, 5], [1, 2, 10]], needs=[3, 2])
+ )
+ print(
+ obj.shoppingOffers(
+ price=[2, 3, 4], special=[[1, 1, 0, 4], [2, 2, 1, 9]], needs=[1, 2, 1]
+ )
+ )