638. Shopping Offers

Problem

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

Solution

我們可以用 backtracking 找出所有的購買組合，然後再取最小的價錢。每一個 special offers 是不限制購買次數的。每次購買一個 special offer 後，我們必須將 needs 中的商品數量減去 special offer 中的商品數量，然後再重複進行一次購買。最後，我們將無法再購買任何 special offer。此時，如果 needs 中還有未購買的數量時，就會用 price 裡的價格。

Backtracking

class Solution {
    public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
        return dfs(price, special, needs, 0);
    }

    private int dfs(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int index) {
        int amount = 0;
        for (int i = 0; i < needs.size(); i++) {
            amount += price.get(i) * needs.get(i);
        }

        for (int i = index; i < special.size(); i++) {
            List<Integer> offer = special.get(i);
            if (!canUseOffer(needs, offer)) continue;

            for (int j = 0; j < needs.size(); j++) {
                needs.set(j, needs.get(j) - offer.get(j));
            }

            amount = Math.min(amount, offer.get(offer.size() - 1) + dfs(price, special, needs, i));

            for (int j = 0; j < needs.size(); j++) {
                needs.set(j, needs.get(j) + offer.get(j));
            }
        }

        return amount;
    }

    private boolean canUseOffer(List<Integer> needs, List<Integer> offer) {
        for (int i = 0; i < needs.size(); i++) {
            if (needs.get(i) < offer.get(i)) return false;
        }
        return true;
    }
}

參考

• 638. Shopping Offers, LeetCode.
• 638. Shopping Offers, LeetCode Solutions.