# Shopping Offers LeetCode Solution

Here, We see Shopping Offers LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

# List of all LeetCode Solution

## Problem Statement

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.

## Shopping Offers LeetCode Solution C++

``````class Solution {
private:
int ex(int ind, vector<int> ds, vector<int> &price, vector<vector<int>> &offer, vector<int> &need,  map<int,map<vector<int>,int>> &dp){
if(ind==offer.size()){
int tot=0;
for(int i=0; i<need.size(); i++){
tot+=(need[i]-ds[i])*price[i];
}
}
if(dp.find(ind)!=dp.end() && dp[ind].find(ds)!=dp[ind].end()){
return dp[ind][ds];
}
int offer_notTake=ex(ind+1,ds,price,offer,need,dp);
int offer_take=1e9;
vector<int> copyDS=ds;
for(int i=0; i<need.size(); i++){
if(copyDS[i]+offer[ind][i]<=need[i]){
copyDS[i]+=offer[ind][i];
}else{
return dp[ind][ds]=min(offer_notTake,offer_take);
}
}
offer_take=offer[ind][price.size()]+ex(ind,copyDS,price,offer,need,dp);
return dp[ind][ds]=min(offer_notTake,offer_take);
}
public:
int shoppingOffers(vector<int>& price, vector<vector<int>>& offer, vector<int>& need) {
map<int,map<vector<int>,int>> dp;
vector<int> ds(need.size(),0);
return ex(0,ds,price,offer,need,dp);
}
};```Code language: PHP (php)```

## Shopping Offers LeetCode Solution Java

``````class Solution {
int min = Integer.MAX_VALUE;
public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
solve(0, price, special, needs, 0);
return min;
}
public void solve(int idx, List<Integer> price, List<List<Integer>> special, List<Integer> needs, int bought) {
if (isFulfilled(needs)) {
min = Math.min(min, bought);
return;
}
if (idx >= special.size()) {
int total = bought;
for (int i = 0; i < needs.size(); i++) {
total += needs.get(i) * price.get(i);
}
min = Math.min(min, total);
return;
}
solve(idx + 1, price, special, needs, bought);
solve(idx, price, special, newNeeds, bought + special.get(idx).get(needs.size()));
}
}
public boolean isFulfilled(List<Integer> needs) {
for (int need : needs) {
if (need != 0) {
return false;
}
}
return true;
}
public boolean canBuy(List<Integer> needs, List<Integer> offer) {
for (int i = 0; i < needs.size(); i++) {
if (needs.get(i) < offer.get(i)) {
return false;
}
}
return true;
}
public List<Integer> buyProduct(List<Integer> needs, List<Integer> offer) {
List<Integer> newNeeds = new ArrayList<>(needs);
for (int i = 0; i < needs.size(); i++) {
newNeeds.set(i, needs.get(i) - offer.get(i));
}
return newNeeds;
}
}```Code language: PHP (php)```

## Shopping Offers LeetCode Solution JavaScript

``````var shoppingOffers = function(price, special, needs) {
let betterSpecials = special.filter((s) =>
!special.some(s2 => s.at(-1) > s2.at(-1) && s.slice(0, -1).every((y, j) => y === s2[j]))
)
function check(price, special, needs){
let offer = needs.reduce((sum, n, i) => sum + n * price[i], 0);
for(const s of special){
if(s.some((x, i) => x > needs[i])) continue;
offer = Math.min(offer, s.at(-1) + check(price, special, needs.map((n, i) => n - s[i])));
}
return offer;
}
return check(price, betterSpecials, needs);
};```Code language: JavaScript (javascript)```

## Shopping Offers Solution Python

``````class Solution(object):
def shoppingOffers(self, price, special, needs):
n = len(price)
def dfs(needs):
ans = sum([i*j for i, j in zip(price, needs)])
cur = sys.maxsize
for s in special:
new_needs, ok = [], True
for i in range(n):
need, give = needs[i], s[i]
if need < give:
ok = False
break
new_needs.append(need-give)
if ok: cur = min(cur, dfs(tuple(new_needs)) + s[-1])
return min(ans, cur)
return dfs(tuple(needs))```Code language: HTML, XML (xml)```
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