Leetcode #2144: Minimum Cost of Buying Candies With Discount
In this guide, we solve Leetcode #2144 Minimum Cost of Buying Candies With Discount 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
A shop is selling candies at a discount. For every two candies sold, the shop gives a third candy for free.
Quick Facts
- Difficulty: Easy
- Premium: No
- Tags: Greedy, Array, Sorting
Intuition
A locally optimal choice leads to a globally optimal result for this structure.
That means we can commit to decisions as we scan without backtracking.
Approach
Sort or preprocess if needed, then repeatedly take the best available local choice.
Maintain the minimal state necessary to validate the greedy decision.
Steps:
- Sort or preprocess as needed.
- Iterate and pick the best local option.
- Track the current solution.
Example
Input: cost = [1,2,3]
Output: 5
Explanation: We buy the candies with costs 2 and 3, and take the candy with cost 1 for free.
The total cost of buying all candies is 2 + 3 = 5. This is the only way we can buy the candies.
Note that we cannot buy candies with costs 1 and 3, and then take the candy with cost 2 for free.
The cost of the free candy has to be less than or equal to the minimum cost of the purchased candies.
Python Solution
class Solution:
def minimumCost(self, cost: List[int]) -> int:
cost.sort(reverse=True)
return sum(cost) - sum(cost[2::3])
Complexity
The time complexity is , and the space complexity is . 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.