How do you solve Buy Two Chocolates on LeetCode?

Buy Two Chocolates (LeetCode #2706) 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: Sorting helps in quickly finding the cheapest options.

What is the brute force solution for Buy Two Chocolates?

The brute force approach for Buy Two Chocolates is Brute Force, with O(n²) time complexity and O(1) space complexity. We can check every possible pair of chocolates to find the two with the lowest combined price. This method is straightforward but inefficient as it checks all combinations. The optimal Optimal Solution approach improves this to O(n log n).

What algorithm or data structure is used to solve Buy Two Chocolates?

Buy Two Chocolates uses the following concepts: Array, Greedy, Sorting. The recommended patterns to study are: Greedy, Sorting.

What are the interview tips for Buy Two Chocolates?

Explain your thought process clearly while coding. Consider edge cases like the minimum and maximum values of input. Practice explaining your solution as if you were teaching someone else.

What are common mistakes when solving Buy Two Chocolates?

Not checking if the sum of the two cheapest chocolates is within budget. Forgetting to return the original money if no valid pair is found.

#2706

Buy Two Chocolates

Easy

ArrayGreedySortingGreedySorting

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 prices first, we can easily find the two cheapest chocolates and check if we can afford them. This significantly reduces the number of checks we need to perform.

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Algorithm

3 steps

1Step 1: Sort the prices array.
2Step 2: Check if the sum of the first two elements (cheapest chocolates) is less than or equal to money.
3Step 3: If yes, return the leftover money after buying them. If no, return the initial money.

solution.py5 lines

1def buyChocolates(prices, money):
2    prices.sort()
3    if prices[0] + prices[1] <= money:
4        return money - (prices[0] + prices[1])
5    return money

ℹ

Complexity note: The time complexity is O(n log n) due to the sorting step, which is more efficient than the O(n²) brute force approach. The space complexity is O(1) since we are not using any additional data structures.

1Sorting helps in quickly finding the cheapest options.
2Always check if you can afford the cheapest two before proceeding.

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