638. Shopping Offers
Description
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 ith
item, and an integer array needs
where needs[i]
is the number of pieces of the ith
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 jth
item in the ith
offer and special[i][n]
(i.e., the last integer in the array) is the price of the ith
offer.
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 than you want, even if that would lower the overall price. You could use any of the special offers as many times as you want.
Example 1:
Input: price = [2,5], special = [[3,0,5],[1,2,10]], needs = [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: price = [2,3,4], special = [[1,1,0,4],[2,2,1,9]], needs = [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.
Constraints:
n == price.length == needs.length
1 <= n <= 6
0 <= price[i], needs[i] <= 10
1 <= special.length <= 100
special[i].length == n + 1
0 <= special[i][j] <= 50
- The input is generated that at least one of
special[i][j]
is non-zero for0 <= j <= n - 1
.
Solutions
Solution 1: State Compression + Memoization Search
We notice that the number of types of items $n \leq 6$ in the problem, and the quantity of each item needed does not exceed $10$. We can use $4$ binary bits to represent the quantity of each item needed. Thus, we only need at most $6 \times 4 = 24$ binary bits to represent the entire shopping list.
First, we convert the shopping list $\textit{needs}$ into an integer $\textit{mask}$, where the quantity of the $i$-th item needed is stored in the $i \times 4$ to $(i + 1) \times 4 - 1$ bits of $\textit{mask}$. For example, when $\textit{needs} = [1, 2, 1]$, we have $\textit{mask} = 0b0001 0010 0001$.
Then, we design a function $\textit{dfs}(cur)$, representing the minimum amount of money we need to spend when the current state of the shopping list is $\textit{cur}$. Therefore, the answer is $\textit{dfs}(\textit{mask})$.
The calculation method of the function $\textit{dfs}(cur)$ is as follows:
- First, we calculate the cost of the current shopping list $\textit{cur}$ without using any bundles, denoted as $\textit{ans}$.
- Then, we iterate through each bundle $\textit{offer}$. If the current shopping list $\textit{cur}$ can use the bundle $\textit{offer}$, i.e., the quantity of each item in $\textit{cur}$ is not less than that in the bundle $\textit{offer}$, then we can try to use this bundle. We subtract the quantity of each item in the bundle $\textit{offer}$ from $\textit{cur}$, obtaining a new shopping list $\textit{nxt}$, then recursively calculate the minimum cost of $\textit{nxt}$ and add the price of the bundle $\textit{offer}[n]$, updating $\textit{ans}$, i.e., $\textit{ans} = \min(\textit{ans}, \textit{offer}[n] + \textit{dfs}(\textit{nxt}))$.
- Finally, return $\textit{ans}$.
To avoid repeated calculations, we use a hash table $\textit{f}$ to record the minimum cost corresponding to each state $\textit{cur}$.
The time complexity is $O(n \times k \times m^n)$, where $n$ represents the types of items, and $k$ and $m$ respectively represent the number of bundles and the maximum demand for each type of item. The space complexity is $O(n \times m^n)$.
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