Leetcode #638: Shopping Offers
In this guide, we solve Leetcode #638 Shopping Offers in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
Problem Statement
In LeetCode Store, there are n items to sell. Each item has a price.
Quick Facts
- Difficulty: Medium
- Premium: No
- Tags: Bit Manipulation, Memoization, Array, Dynamic Programming, Backtracking, Bitmask
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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.
Python Solution
from functools import lru_cache
from typing import List
def shoppingOffers(price: List[int], special: List[List[int]], needs: List[int]) -> int:
n = len(price)
filtered = []
for sp in special:
cost = sp[-1]
if cost < sum(sp[i] * price[i] for i in range(n)):
filtered.append(sp)
def dfs(state):
state = list(state)
best = sum(state[i] * price[i] for i in range(n))
for sp in filtered:
nxt = []
for i in range(n):
if sp[i] > state[i]:
break
nxt.append(state[i] - sp[i])
else:
best = min(best, sp[-1] + dfs(tuple(nxt)))
return best
return dfs(tuple(needs))
Complexity
The time complexity is , where represents the types of items, and and respectively represent the number of bundles and the maximum demand for each type of item. The space complexity is .
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.