Leetcode 2706. Buy Two Chocolates Sure, let’s tackle the problem step-by-step.
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.
To ensure we fully understand the problem, we should clarify a few points:
Assuming:
To solve this problem:
money
.money
since it’s not possible to buy two chocolates.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
}
}
prices
array takes O(n log n) time, where n
is the number of elements in the array.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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