LeetCode 638. Shopping Offers

原题链接在这里：https://leetcode.com/problems/shopping-offers/description/

题目：

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.

题解：

For recursion, it needs to parameter to calculate current state.

For current state, it needs price and current needs to calcuate direct purchase price.

Then for each offer in special, check if it is valid, if not, skip. If yes, minius offer items in needs and do next level.

This question is bottom up dfs. return the minimum value for current needs.

Time Complexity: O(special.size()*needs.size()*n), n 是recursive call的层数, 可以是needs中所有integer最大的值.

Space: O(n).

AC Java:

class Solution {
    public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
        return dfs(price, special, needs);
    }
    
    private int dfs(List<Integer> price, List<List<Integer>> special, List<Integer> needs){
        // Direct purchase price
        int res = directPurchase(price, needs);
        for(List<Integer> offer : special){
            if(!isValid(offer, needs)){
                continue;
            }
            
            List<Integer> restNeeds = new ArrayList<Integer>();
            for(int i = 0; i<needs.size(); i++){
                restNeeds.add(needs.get(i)-offer.get(i));
            }
            
            res = Math.min(res, offer.get(offer.size()-1) + dfs(price, special, restNeeds));
        }
        
        return res;
    }
    
    
    private int directPurchase(List<Integer> price, List<Integer> needs){
        int res = 0;
        for(int i = 0; i<price.size(); i++){
            res += price.get(i)*needs.get(i);
        }
        
        return res;
    }
    
    private boolean isValid(List<Integer> offer, List<Integer> needs){
        for(int i = 0; i<needs.size(); i++){
            if(offer.get(i) > needs.get(i)){
                return false;
            }
        }
        
        return true;
    }
}

类似Combination Sum.