Sort the array of prices and use a two-pointer technique to find the two cheapest chocolates that can be bought such that their sum is less than or equal to the given amount of money. This approach helps efficiently find a solution by reducing the problem space through sorting.
Time Complexity: O(n log n) due to sorting.
Space Complexity: O(1) as the sorting is in-place.
1using System;
2
3class Program {
4 public static int BuyTwoChocolates(int[] prices, int money) {
5 Array.Sort(prices);
6 int left = 0, right = prices.Length - 1;
7 int minSpent = int.MaxValue;
8
9 while (left < right) {
10 int sum = prices[left] + prices[right];
11 if (sum <= money) {
12 minSpent = sum;
13 left++;
14 } else {
15 right--;
16 }
17 }
18 return (minSpent == int.MaxValue) ? money : money - minSpent;
19 }
20
21 static void Main() {
22 int[] prices = { 1, 2, 2 };
23 int money = 3;
24 Console.WriteLine(BuyTwoChocolates(prices, money));
25 }
26}The C# approach follows similar logic by sorting the array and using two pointers. The result is the original money value minus the cost of the selected chocolates.
In this approach, check all possible pairs of chocolate prices. If a pair's sum is within the available money and minimal, update the result. Although this approach is less efficient, it is a simple and direct way to solve the problem given the constraints.
Time Complexity: O(n^2) due to the nested loop.
Space Complexity: O(1) with a few extra variables.
1#include <iostream>
2#include <vector>
3#include <climits>
4
5int buyTwoChocolates(std::vector<int>& prices, int money) {
6 int minSpent = money + 1;
7 for (size_t i = 0; i < prices.size(); ++i) {
8 for (size_t j = i + 1; j < prices.size(); ++j) {
9 int sum = prices[i] + prices[j];
10 if (sum <= money && sum < minSpent) {
11 minSpent = sum;
12 }
13 }
14 }
15 return (minSpent > money) ? money : money - minSpent;
16}
17
18int main() {
19 std::vector<int> prices = {3, 2, 3};
20 int money = 3;
21 std::cout << buyTwoChocolates(prices, money) << std::endl;
22 return 0;
23}The brute force method involves exploring each pair via nested loops, updating the minimal sum and ensuring the selected pair does not overspend the allocated money.