A store offers a number of special deals, and they would like to offer customers optimal conditions. The store holds several products, and each product has a fixed price. A customer can buy products in standard quantities at the regular price or via special offers. Each special offer bundles certain quantities of products at a potentially lower price.
You are given:
price
where price[i]
is the price of the i-th product.needs
where needs[i]
is the number of units of the i-th product that you need to buy.special
offers, where each special[i]
is a list with the last element being the price of the special offer and the rest of the elements represent the quantity of each product included in the offer.You need to return the minimum cost to satisfy the shopping list needs
.
needs
list be zero?
price
, needs
, or special
lists?
<= 6
for products and may vary accordingly for special offers.needs
?
To solve this problem, we can use a Depth-First Search (DFS) approach with memoization to reduce the computational overlap. Here are the primary steps:
import java.util.HashMap;
import java.util.List;
import java.util.Map;
public class Solution {
// Memoization map
private Map<List<Integer>, Integer> memo = new HashMap<>();
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) {
if (memo.containsKey(needs)) {
return memo.get(needs);
}
// Calculate the direct cost without any offers
int minCost = 0;
for (int i = 0; i < needs.size(); i++) {
minCost += needs.get(i) * price.get(i);
}
// Try every special offer
for (List<Integer> offer : special) {
List<Integer> cloneNeeds = needs.clone();
boolean valid = true;
for (int i = 0; i < needs.size(); i++) {
int diff = cloneNeeds.get(i) - offer.get(i);
if (diff < 0) {
valid = false;
break;
}
cloneNeeds.set(i, diff);
}
if (valid) {
int offerCost = offer.get(offer.size() - 1) + dfs(price, special, cloneNeeds);
minCost = Math.min(minCost, offerCost);
}
}
memo.put(needs, minCost);
return minCost;
}
}
The time complexity is challenging to pinpoint exactly due to the recursion and memoization. In the worst-case scenario, each need for each product can range from 0 to some maximum value k
. The theoretical upper bound could be exponential in terms of the number of products (O(k^n)
) but is significantly mitigated by memoization, which avoids solving the same subproblem multiple times.
With memoization, the practical performance is much better and depends on the number of unique states the needs
list can represent and memoize.
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