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 $2and$5 respectively. In special offer 1, you can pay $5for3Aand0BInspecialoffer2,youcanpay$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, $4forC.Youmaypay$4 for 1A and 1B, and $9for2A,2Band1C.Youneedtobuy1A,2Band1C,soyoumaypay$4 for 1A and 1B (special offer #1), and $3for1B,$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

Code

Time Complexity: $O(n^2)$, Space Complexity: $O(n^2)$

Intuition 就是：special offer 每個都只有兩種選擇，用或不用。就是 knapsack 的概念。

class Solution {
    map<vector<int>, int> memo;
public:
    int shoppingOffers(vector<int>& price, vector<vector<int>>& special, vector<int>& needs) {
        // speed up
        if(memo[needs]) return memo[needs];
 
        int bestPrice = calculatePrice(price, needs);
        for(auto& sp: special) {
            useOffer(sp, needs);
            if(noNegatives(needs)) {
                int priceWithOffer = sp.back() + shoppingOffers(price, special, needs);
                bestPrice = min(bestPrice, priceWithOffer);
            }
            restoreOffer(sp, needs);
        }
        memo[needs] = bestPrice;
        return bestPrice;
    }
 
    int calculatePrice(vector<int>& price, vector<int>& needs) {
        int total = 0;
        for(int i = 0; i < price.size(); i++) {
            total += price[i] * needs[i];
        }
        return total;
    }
 
    void useOffer(vector<int>& offer, vector<int>& needs){
        for(int i = 0; i < needs.size(); i++) {
            needs[i] -= offer[i];
        }
    }
 
    void restoreOffer(vector<int>& offer, vector<int>& needs){
        for(int i = 0; i < needs.size(); i++) {
            needs[i] += offer[i];
        }
    }
 
    bool noNegatives(vector<int>& needs) {
        for(auto n: needs) {
            if(n < 0) return false;
        }
        return true;
    }
 
};

Source

• Shopping Offers - LeetCode