638. Shopping Offers

1. Question

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.

2. Examples

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.

3. Constraints

• n == price.length
• n == needs.length
• 1 <= n <= 6
• 0 <= price[i] <= 10
• 0 <= needs[i] <= 10
• 1 <= special.length <= 100
• special[i].length == n + 1
• 0 <= special[i][j] <= 50

4. References

来源：力扣（LeetCode） 链接：https://leetcode-cn.com/problems/shopping-offers 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

5. Solutions

不会做，参考大神的，加了注释。

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

  // index为special尝试的下标索引
  public int shopping(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int index) {
    // 大礼包遍历结束，尝试原价购买所有东西
    if (index == special.size()) {
      return purchaseWithOriginalPrice(price, needs);
    }

    // 将需要购买的清单拷贝
    List<Integer> clone = new ArrayList<>(needs);
    int i = 0;
    // 遍历指定index的大礼包中的各项
    for (; i < special.get(index).size() - 1; i++) {
      // 对大礼包中各项进行相减
      int remain = needs.get(i) - special.get(index).get(i);
      // 东西买多了，排除这项可能
      if (remain < 0) {
        break;
      } else {
        // 如果有剩余东西没有购买，符合条件，修改购物清单
        clone.set(i, remain);
      }
    }

    // 遍历上述大礼包结束，进行递归，查看是继续使用大礼包便宜还是原价购买便宜
    // i正好是special处对应的价格的索引
    if (i == special.get(index).size() - 1) {
      // 值得注意的是，大礼包可以重复使用
      return Math.min(special.get(index).get(i) + shopping(price, special, clone, index), shopping(price, special, needs, index + 1));            
    } else {
      // 如果i不能到special中对应的价格的索引，证明不能使用大礼包。直接按照原价购买
      return shopping(price, special, needs, index + 1);
    }
  }

  // 按照原价购买
  public int purchaseWithOriginalPrice(List<Integer> price, List<Integer> needs) {
    int sum = 0;
    for (int i = 0; i < needs.size(); i++) {
      sum += needs.get(i) * price.get(i);
    }
    return sum;
  }    
}