How do you solve Minimum Cost of Buying Candies With Discount on LeetCode?

Minimum Cost of Buying Candies With Discount (LeetCode #2144) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n log n) time complexity and O(1) space complexity. Key insight: Buying the most expensive candies first maximizes the value of the free candy.

What is the brute force solution for Minimum Cost of Buying Candies With Discount?

The brute force approach for Minimum Cost of Buying Candies With Discount is Brute Force, with O(n²) time complexity and O(1) space complexity. In the brute-force approach, we consider every possible combination of buying two candies and then check which candy can be taken for free. This method is straightforward but inefficient as it checks all pairs. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Minimum Cost of Buying Candies With Discount?

Minimum Cost of Buying Candies With Discount uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy Algorithms, Sorting.

What are the interview tips for Minimum Cost of Buying Candies With Discount?

Always think about how to maximize or minimize values in problems involving costs. Discuss your thought process out loud; it shows your reasoning and can lead to hints from the interviewer. Practice sorting and greedy algorithms, as they are common in optimization problems.

What are common mistakes when solving Minimum Cost of Buying Candies With Discount?

Not considering the free candy's eligibility based on the minimum cost of bought candies. Failing to sort the costs before calculating the total, which can lead to suboptimal solutions.

#2144

Minimum Cost of Buying Candies With Discount

Easy

ArrayGreedySortingGreedy AlgorithmsSorting

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n log n)
Space	O(1)	O(1)

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Intuition

Time O(n log n)Space O(1)

By sorting the candy costs in descending order, we can maximize the value of the free candy we get. We buy the most expensive candies first and always take the next cheapest candy that is eligible for free.

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Algorithm

5 steps

1Step 1: Sort the candies in descending order of cost.
2Step 2: Initialize a total cost variable to 0.
3Step 3: Iterate through the sorted list, buying two candies at a time.
4Step 4: For every two candies bought, skip the next candy (which is the free one).
5Step 5: Add the costs of the bought candies to the total cost.

solution.py7 lines

1def minimumCost(cost):
2    cost.sort(reverse=True)
3    total_cost = 0
4    for i in range(len(cost)):
5        if (i % 3) != 2:
6            total_cost += cost[i]
7    return total_cost

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, which is the most time-consuming operation. The subsequent iteration through the list is O(n), making the overall complexity dominated by the sorting step.

1Buying the most expensive candies first maximizes the value of the free candy.
2Sorting the costs allows for a structured approach to minimize total spending.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.