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 i^{th}
item, and an integer array needs
where needs[i]
is the number of pieces of 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}
offer and special[i][n]
(i.e., the last integer in the array) is the price of the i^{th}
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
Solutions#
Solution 1#
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 class Solution:
def shoppingOffers(
self, price: List[int], special: List[List[int]], needs: List[int]
) > int:
def total(price, needs):
return sum(price[i] * needs[i] for i in range(len(needs)))
ans = total(price, needs)
t = []
for offer in special:
t.clear()
for j in range(len(needs)):
if offer[j] > needs[j]:
t.clear()
break
t.append(needs[j]  offer[j])
if t:
ans = min(ans, offer[1] + self.shoppingOffers(price, special, t))
return ans

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 class Solution {
public int shoppingOffers(
List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
int ans = total(price, needs);
List<Integer> t = new ArrayList<>();
for (List<Integer> offer : special) {
t.clear();
for (int j = 0; j < needs.size(); ++j) {
if (offer.get(j) > needs.get(j)) {
t.clear();
break;
}
t.add(needs.get(j)  offer.get(j));
}
if (!t.isEmpty()) {
ans = Math.min(
ans, offer.get(offer.size()  1) + shoppingOffers(price, special, t));
}
}
return ans;
}
private int total(List<Integer> price, List<Integer> needs) {
int s = 0;
for (int i = 0; i < price.size(); ++i) {
s += price.get(i) * needs.get(i);
}
return s;
}
}

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 class Solution {
public:
int shoppingOffers(vector<int>& price, vector<vector<int>>& special, vector<int>& needs) {
int ans = total(price, needs);
vector<int> t;
for (auto& offer : special) {
t.clear();
for (int j = 0; j < needs.size(); ++j) {
if (offer[j] > needs[j]) {
t.clear();
break;
}
t.push_back(needs[j]  offer[j]);
}
if (!t.empty()) ans = min(ans, offer[offer.size()  1] + shoppingOffers(price, special, t));
}
return ans;
}
int total(vector<int>& price, vector<int>& needs) {
int s = 0;
for (int i = 0; i < price.size(); ++i) s += price[i] * needs[i];
return s;
}
};

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 func shoppingOffers(price []int, special [][]int, needs []int) int {
total := func(price, needs []int) int {
s := 0
for i := 0; i < len(needs); i++ {
s += price[i] * needs[i]
}
return s
}
min := func(a, b int) int {
if a < b {
return a
}
return b
}
ans := total(price, needs)
var t []int
for _, offer := range special {
t = t[:0]
for j := 0; j < len(needs); j++ {
if offer[j] > needs[j] {
t = t[:0]
break
}
t = append(t, needs[j]offer[j])
}
if len(t) > 0 {
ans = min(ans, offer[len(offer)1]+shoppingOffers(price, special, t))
}
}
return ans
}
