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Java-based LeetCode algorithm problem solutions, regularly updated
package g0601_0700.s0638_shopping_offers;
// #Medium #Array #Dynamic_Programming #Bit_Manipulation #Backtracking #Bitmask #Memoization
// #2022_03_21_Time_6_ms_(71.06%)_Space_42.6_MB_(71.63%)
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
/**
* 638 - Shopping Offers\.
*
* Medium
*
* 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`
* * `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`
**/
public class Solution {
public int shoppingOffers(
List price, List> special, List needs) {
Map, Integer> map = new HashMap<>();
shoppingOffersUtil(price, special, needs, map);
return map.get(needs);
}
private int shoppingOffersUtil(
List price,
List> special,
List needs,
Map, Integer> map) {
if (map.containsKey(needs)) {
return map.get(needs);
}
Integer ans = computePrice(price, needs);
for (int i = 0; i < special.size(); i++) {
if (verify(special.get(i), needs)) {
ans =
Math.min(
special.get(i).get(needs.size())
+ shoppingOffersUtil(
price,
special,
updatedNeeds(needs, special.get(i)),
map),
ans);
}
}
map.put(needs, ans);
return map.get(needs);
}
private List updatedNeeds(List needs, List special) {
List updatedNeeds = new ArrayList<>(needs);
for (int i = 0; i < needs.size(); i++) {
updatedNeeds.set(i, updatedNeeds.get(i) - special.get(i));
}
return updatedNeeds;
}
private boolean verify(List special, List needs) {
for (int i = 0; i < needs.size(); i++) {
if (special.get(i) > needs.get(i)) {
return false;
}
}
return true;
}
private Integer computePrice(List price, List needs) {
int ans = 0;
for (int i = 0; i < needs.size(); i++) {
ans += (needs.get(i) * price.get(i));
}
return ans;
}
}