638. Shopping Offers LeetCode Solution

Last updated: February 19, 2025

In this guide, you will get 638. Shopping Offers LeetCode Solution with the best time and space complexity. The solution to Shopping Offers problem is provided in various programming languages like C++, Java, and Python. This will be helpful for you if you are preparing for placements, hackathons, interviews, or practice purposes. The solutions provided here are very easy to follow and include detailed explanations.

Problem Statement
Complexity Analysis
Shopping Offers solution in C++
Shopping Offers solution in Java
Shopping Offers solution in Python
Additional Resources

638. Shopping Offers LeetCode Solution image

Problem Statement of Shopping Offers

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 == needs.length
1 <= n <= 6
0 <= price[i], needs[i] <= 10
1 <= special.length <= 100
special[i].length == n + 1
0 <= special[i][j] <= 50
The input is generated that at least one of special[i][j] is non-zero for 0 <= j <= n – 1.

Complexity Analysis

Time Complexity: O(|\texttt{special}||\texttt{needs}|k), where k = \text{max of needs} = 6
Space Complexity: O(6)

638. Shopping Offers LeetCode Solution in C++

class Solution {
 public:
  int shoppingOffers(vector<int>& price, vector<vector<int>>& special,
                     vector<int>& needs) {
    return dfs(price, special, needs, 0);
  }

 private:
  int dfs(const vector<int>& price, const vector<vector<int>>& special,
          vector<int>& needs, int s) {
    int ans = 0;
    for (int i = 0; i < price.size(); ++i)
      ans += price[i] * needs[i];

    for (int i = s; i < special.size(); ++i)
      if (isValid(special[i], needs)) {
        // Use the special[i].
        for (int j = 0; j < needs.size(); ++j)
          needs[j] -= special[i][j];
        ans = min(ans, special[i].back() + dfs(price, special, needs, i));
        // Unuse the special[i] (backtracking).
        for (int j = 0; j < needs.size(); ++j)
          needs[j] += special[i][j];
      }

    return ans;
  }

  // Returns true if this special offer is a valid one.
  bool isValid(const vector<int>& offer, const vector<int>& needs) {
    for (int i = 0; i < needs.size(); ++i)
      if (needs[i] < offer[i])
        return false;
    return true;
  }
};
/* code provided by PROGIEZ */

638. Shopping Offers LeetCode Solution in Java

class Solution {
  public int shoppingOffers(List<Integer> price, List<List<Integer>> special, List<Integer> needs) {
    return dfs(price, special, needs, 0);
  }

  private int dfs(List<Integer> price, List<List<Integer>> special, List<Integer> needs, int s) {
    int ans = 0;
    for (int i = 0; i < needs.size(); ++i)
      ans += needs.get(i) * price.get(i);

    for (int i = s; i < special.size(); ++i) {
      List<Integer> offer = special.get(i);
      if (isValid(offer, needs)) {
        // Use the special[i].
        for (int j = 0; j < needs.size(); ++j)
          needs.set(j, needs.get(j) - offer.get(j));
        ans = Math.min(ans, offer.get(offer.size() - 1) + dfs(price, special, needs, i));
        // Unuse the special[i] (backtracking).
        for (int j = 0; j < needs.size(); ++j)
          needs.set(j, needs.get(j) + offer.get(j));
      }
    }

    return ans;
  }

  // Returns true if this special offer is a valid one.
  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;
  }
}
// code provided by PROGIEZ

638. Shopping Offers LeetCode Solution in Python

class Solution:
  def shoppingOffers(
      self,
      price: list[int],
      special: list[list[int]],
      needs: list[int]
  ) -> int:
    def dfs(s: int) -> int:
      ans = 0
      for i, need in enumerate(needs):
        ans += need * price[i]

      for i in range(s, len(special)):
        offer = special[i]
        if all(offer[j] <= need for j, need in enumerate(needs)):
          # Use the special[i].
          for j in range(len(needs)):
            needs[j] -= offer[j]
          ans = min(ans, offer[-1] + dfs(i))
          # Unuse the special[i] (backtracking).
          for j in range(len(needs)):
            needs[j] += offer[j]

      return ans

    return dfs(0)
 # code by PROGIEZ

Additional Resources

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