LC 638. Shopping Offers

In LeetCode Store, there are some kinds of 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 the each item's price, a set of special offers, and the number we need to buy for each item. The job is to output the lowest price you have to pay for exactly certain items as given, where you could make optimal use of the special offers.

Each special offer is represented in the form of an array, the last number represents the price you need to pay for this special offer, other numbers represents how many specific items you could get if you buy this offer.

You could use any of special offers as many times as you want.

Example 1:

Input: [2,5], [[3,0,5],[1,2,10]], [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: [2,3,4], [[1,1,0,4],[2,2,1,9]], [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.

Note:

1. There are at most 6 kinds of items, 100 special offers.
2. For each item, you need to buy at most 6 of them.
3. You are not allowed to buy more items than you want, even if that would lower the overall price.

Runtime: 53 ms, faster than 12.62% of Java online submissions for Shopping Offers.

自己的揭发，但是时间有点低，纯粹暴力搜索。

先把总的不加优惠券的价格算出来，然后每次遍历优惠券，直到不能再用优惠券。每次用优惠券都是DFS，且保存当前优化的结果，这又是BFS。

class Solution {
  public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
    int maxval = 0;
    int cnt = 0;
    for(int x : needs){
      maxval += x * price.get(cnt);
      cnt++;
    }
    return helper(price, special, needs, maxval);
  }
  int helper(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int maxval){
    int tmpmax = maxval;
    for(int i=0; i<special.size(); i++){
      List<Integer> newneeds = new ArrayList<>();
      for(int n : needs) newneeds.add(n);
      boolean canuse = true;
      int delta = maxval;
      for(int j=0; j<special.get(i).size()-1; j++){
        if(newneeds.get(j) < special.get(i).get(j)){
          canuse = false;
          break;
        }else {
          delta -= special.get(i).get(j) * price.get(j);
          newneeds.set(j, newneeds.get(j) - special.get(i).get(j));
        }
      }
      if(canuse) {
        delta += special.get(i).get(special.get(i).size()-1);
        tmpmax = Math.min(tmpmax, helper(price, special, newneeds, delta));
      }
    }
    return tmpmax;
  }
}

一个更好的解法

Runtime: 12 ms, faster than 90.29% of Java online submissions for Shopping Offers.

优化解法，

1. 遍历的时候，可以记录当前index，也就是说不需要每次从头遍历，因为只是使用优惠券顺序的关系而已。

2. 我之前会用backtracking偷懒只用一个needs，用到了set，这十分费时。直接开一个新的newneeds，每次用add，效果十分好。

class Solution {
  public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
    int maxval = 0;
    int cnt = 0;
    for(int x : needs){
      maxval += x * price.get(cnt);
      cnt++;
    }
    return helper(price, special, needs, maxval,0);
  }
  int helper(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int maxval, int spindex){
    int tmpmax = maxval;
//    StringBuilder sb = new StringBuilder();
//    for(int i=0; i<needs.size(); i++){
//      sb.append(needs.get(i));
//      sb.append("+");
//    }
//    if(mp.containsKey(sb.toString())) return mp.get(sb.toString());
    for(int i=spindex; i<special.size(); i++){
      //List<Integer> newneeds = new ArrayList<>();
      //for(int n : needs) newneeds.add(n);
      //if(special.get(i).get(special.get(i).size()-1) <= maxval - tmpmax) continue;
      //boolean canuse = true;
      int delta = maxval;
      List<Integer> newneeds = new ArrayList<>();
      for(int j=0; j<special.get(i).size()-1; j++){
        //if(newneeds.get(j) < special.get(i).get(j)){
        if(needs.get(j) < special.get(i).get(j)){
          //canuse = false;
          newneeds = null;
          break;
          //needs.set(j, needs.get(j) - special.get(i).get(j));
        }else {
          delta -= special.get(i).get(j) * price.get(j);
          newneeds.add(needs.get(j) - special.get(i).get(j));
          //newneeds.set(j, newneeds.get(j) - special.get(i).get(j));
          //needs.set(j, needs.get(j) - special.get(i).get(j));
        }
      }
      if(newneeds != null) {
        delta += special.get(i).get(special.get(i).size()-1);
        tmpmax = Math.min(tmpmax, helper(price, special, newneeds, delta, i));
      }
      // for(int k=0; k<special.get(i).size()-1; k++){
      //   needs.set(k, needs.get(k) + special.get(i).get(k));
      // }
    }
//    sb = new StringBuilder();
//    for(int k=0; k<needs.size(); k++){
//      sb.append(k);
//      sb.append("+");
//    }
//    mp.put(sb.toString(), maxval);
    return tmpmax;
  }
}