Buy Two Chocolates Interview Question for Google

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.

For example:

If prices = [1, 2, 2] and money = 3, the output should be 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.

If prices = [3, 2, 3] and money = 3, the output should be 3. Explanation: You cannot buy 2 chocolates without going in debt, so we return 3.

Naive Solution

The simplest approach is to iterate through all possible pairs of chocolates, calculate the sum of their prices, and check if buying them leaves a non-negative amount of money. We keep track of the minimum sum that satisfies this condition.

Code

def solve_naive(prices, money):
    min_cost = float('inf')
    for i in range(len(prices)):
        for j in range(i + 1, len(prices)):
            cost = prices[i] + prices[j]
            if money - cost >= 0:
                min_cost = min(min_cost, cost)
    if min_cost == float('inf'):
        return money
    else:
        return money - min_cost

Time Complexity

O(n^2), where n is the number of chocolates (length of prices array), because of the nested loops.

Space Complexity

O(1), as we are only using a constant amount of extra space.

Optimal Solution

To optimize, we can find the two cheapest chocolates. Sort the array, and the first two elements are the cheapest.

Code

def solve_optimal(prices, money):
    prices.sort()
    cost = prices[0] + prices[1]
    if money - cost >= 0:
        return money - cost
    else:
        return money

Time Complexity

O(n log n) due to the sorting of the prices array.

Space Complexity

O(1) or O(n) depending on the sorting algorithm implementation. Some sorting algorithms like quicksort can take O(log n) space, while others like merge sort can take O(n) space.

Edge Cases

Not enough money: If the sum of the two cheapest chocolates is more than the money, return the original money.
Sufficient money: If there's enough money to buy the two cheapest, return the remaining money.
Prices array has only two elements: The algorithm correctly handles this as it directly considers these two.