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.

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: Dynamic programming

Bit manipulation

Data structures

Complete search

Starter code: class Solution:
    def shoppingOffers(self, price: List[int], special: List[List[int]], needs: List[int]) -> int:
        

Solutions:
class Solution:

	def shoppingOffers(self, price, special, needs):

		def dfs(curr, special, needs):
			p = curr + sum((p * needs[i] for (i, p) in enumerate(price)))
			for si in range(len(special)):
				s = special[si]
				if all((n >= s[i] for (i, n) in enumerate(needs))):
					p = min(p, dfs(curr + s[-1], special[si:], [n - s[i] for (i, n) in enumerate(needs)]))
			return p
		return dfs(0, special, needs)


class Solution:

	def shoppingOffers(self, price, special, needs):

		def directPurchase(price, needs):
			total = 0
			for i in range(len(needs)):
				total += price[i] * needs[i]
			return total

		def dfs(price, special, needs, index):
			local_min = directPurchase(price, needs)
			for i in range(index, len(special)):
				offer = special[i]
				temp = []
				for j in range(len(needs)):
					if needs[j] < offer[j]:
						temp = []
						break
					else:
						temp.append(needs[j] - offer[j])
				if temp != []:
					local_min = min(local_min, offer[-1] + dfs(price, special, temp, i))
			return local_min
		return dfs(price, special, needs, 0)


class Solution:

	def shoppingOffers(self, price, special, needs):
		d = {}
		n = len(needs)

		def minp(needs, d):
			if str(needs) in d:
				return d[str(needs)]
			res = 0
			for i in range(n):
				res += price[i] * needs[i]
			for s in special:
				for i in range(n):
					if s[i] > needs[i]:
						break
				else:
					need2 = []
					for i in range(n):
						need2.append(needs[i] - s[i])
					res = min(res, minp(need2, d) + s[-1])
			d[str(needs)] = res
			return res
		return minp(needs, d)


class Solution:

	def shoppingOffers(self, rps, sos, tq):
		if len(sos) < 1:
			return sum([x * y for (x, y) in zip(rps, tq)])
		so = sos.pop()
		m = self.shoppingOffers(rps, sos, tq)
		ss = so[:]
		while all([x <= y for (x, y) in zip(ss, tq)]):
			rq = [x - y for (x, y) in zip(tq, ss)]
			m = min(m, ss[-1] + self.shoppingOffers(rps, sos, rq))
			ss = [x + y for (x, y) in zip(ss, so)]
		sos.append(so)
		return m


class Solution:

	def shoppingOffers(self, price, special, needs):
		res = 0
		for (pric, need) in zip(price, needs):
			res += pric * need
		for offer in special:
			clone = list(needs)
			i = 0
			while i < len(needs):
				diff = clone[i] - offer[i]
				if diff < 0:
					break
				clone[i] = diff
				i += 1
			if i == len(needs):
				res = min(res, offer[-1] + self.shoppingOffers(price, special, clone))
		return res


class Solution:

	def shoppingOffers(self, price, special, needs):
		a = 0
		n = len(needs)
		for i in range(n):
			a += needs[i] * price[i]
		for s in special:
			new_needs = needs.copy()
			check = True
			for i in range(n):
				new_needs[i] = new_needs[i] - s[i]
				if new_needs[i] < 0:
					check = False
					break
			if check:
				a = min(a, s[n] + self.shoppingOffers(price, special, new_needs))
		return a