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Leetcode 2706. Buy Two Chocolates Sure, let’s tackle the problem step-by-step.

Problem Statement

You are given an integer array prices representing the prices of various chocolates in a store, and an integer money representing the amount of money you have. You need to determine the maximum amount of money you can have left after buying exactly two chocolates, or return the total money if it’s not possible to buy two chocolates.

Clarifying Questions

To ensure we fully understand the problem, we should clarify a few points:

  1. Are the prices and money guaranteed to be non-negative integers?
  2. Is the length of the prices array guaranteed to be at least 2? If not, what should be the behavior if it has fewer than two elements?
  3. Can we assume that all prices are unique, or are duplicates allowed?

Assuming:

  1. Yes, prices and money are non-negative integers.
  2. We handle cases where there are fewer than two elements.
  3. Duplicates are allowed.

Strategy

To solve this problem:

  1. Sort the prices array. This allows us to easily find the two cheapest chocolates.
  2. Check if the sum of the two cheapest chocolates is less than or equal to money.
  3. If yes, calculate the remaining money after buying these two chocolates and return it.
  4. If no, return the total money since it’s not possible to buy two chocolates.

Code

Here is the implementation in Java:

import java.util.Arrays;

public class BuyTwoChocolates {

    public static int buyTwoChocolates(int[] prices, int money) {
        // If there are fewer than 2 chocolates, we can't buy two
        if (prices.length < 2) {
            return money;
        }
        
        // Sort the array to get the two cheapest chocolates
        Arrays.sort(prices);
        
        // Sum of the cheapest two chocolates
        int sumOfTwoCheapest = prices[0] + prices[1];
        
        // If the sum is within the available money, return remaining money
        if (sumOfTwoCheapest <= money) {
            return money - sumOfTwoCheapest;
        }
        
        // If not enough money to buy the two cheapest, return the total money
        return money;
    }

    public static void main(String[] args) {
        int[] prices = { 1, 2, 3 };
        int money = 5;
        System.out.println(buyTwoChocolates(prices, money));  // Output: 2
    }
}

Time Complexity

The space complexity is O(1) additional space, assuming the sort operation is done in place (or O(n) if the sorting algorithm creates a copy of the array).

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