Buy Two Chocolates

Problem Description

You are given an integer array prices representing the prices of various chocolates in a store. You are also given a single integer money, which represents your initial amount of money.

You must buy exactly two chocolates in such a way that you still have some non-negative leftover money. You would like to minimize the sum of the prices of the two chocolates you buy.

Return the amount of money you will have leftover after buying the two chocolates. If there is no way for you to buy two chocolates without ending up in debt, return money. Note that the leftover must be non-negative.

Example

Example 1:

Input: prices = [1,2,2], money = 3
Output: 0
Explanation: Purchase the chocolates priced at 1 and 2 units respectively. You will have 3 - 3 = 0 units of money afterwards. Thus, we return 0.

Example 2:

Input: prices = [3,2,3], money = 3
Output: 3
Explanation: You cannot buy 2 chocolates without going in debt, so we return 3.

Constraints

• 2 <= prices.length <= 50

Solution Approach

Intuition:

To efficiently determine the amount of money you will have leftover after buying the two chocolates

Solution Implementation

Code In Different Languages:

JavaScript

Written by @Ishitamukherjee2004

class Solution {
buyChoco(prices, money) {
 prices.sort((a, b) => a - b);
 if (prices[0] + prices[1] > money) {
   return money;
 } else {
   return money - prices[0] - prices[1];
 }
}
}

TypeScript

Written by @Ishitamukherjee2004

 class Solution {
buyChoco(prices: number[], money: number): number {
 prices.sort((a, b) => a - b);
 if (prices[0] + prices[1] > money) {
   return money;
 } else {
   return money - prices[0] - prices[1];
 }
}
}

Python

Written by @Ishitamukherjee2004

 
class Solution:
def buyChoco(self, prices: List[int], money: int) -> int:
 prices.sort()
 if prices[0] + prices[1] > money:
   return money
 else:
   return money - prices[0] - prices[1]

Java

Written by @Ishitamukherjee2004

 public class Solution {
public int buyChoco(int[] prices, int money) {
 Arrays.sort(prices);
 if (prices[0] + prices[1] > money) {
   return money;
 } else {
   return money - prices[0] - prices[1];
 }
}
}

C++

Written by @Ishitamukherjee2004

class Solution {
public:
 int buyChoco(vector<int>& prices, int money) {
     sort(prices.begin(), prices.end());
     if(prices[0]+prices[1]>money) return money;
     else return money-prices[0]-prices[1];
 }
};

Complexity Analysis

• Time Complexity: $O(n*log(n))$
• Space Complexity: $O(1)$
• The time complexity is $O(n*log(n))$ where n is the length of the prices array. This is because the function sorts the prices array.
• The space complexity is $O(1)$ because we are not using any extra space.