Leetcode 2706: Buy Two Chocolates
You are given an integer array representing the prices of chocolates in a store and a total amount of money you have. Your task is to buy exactly two chocolates in a way that ensures you still have some leftover money, and you want to minimize the sum of the prices of the two chocolates. If you cannot buy two chocolates without going into debt, return the original amount of money you had.
Problem
Approach
Steps
Complexity
Input: You are provided with a list of chocolate prices and an integer representing your total amount of money.
Example: Input: prices = [2, 3, 5], money = 6
Constraints:
• 2 <= prices.length <= 50
• 1 <= prices[i] <= 100
• 1 <= money <= 100
Output: The output should be the amount of money you will have left after buying two chocolates, or the original amount of money if it is not possible to buy two chocolates.
Example: Output: 0
Constraints:
• The leftover money should be non-negative.
Goal: The goal is to find the least sum of prices for two chocolates that fit within the available budget.
Steps:
• Step 1: Identify the two cheapest chocolates in the list of prices.
• Step 2: Check if the sum of the prices of these two chocolates is less than or equal to the available money.
• Step 3: If so, return the leftover money after purchasing the two chocolates.
• Step 4: If not, return the original amount of money.
Goal: The prices of the chocolates and the available money should be considered in such a way that you don't end up in debt.
Steps:
• The number of chocolate prices is between 2 and 50.
• Each chocolate price is between 1 and 100.
• The money you have is between 1 and 100.
Assumptions:
• You can buy exactly two chocolates from the store.
• Input: Input: prices = [1, 2, 3], money = 4
• Explanation: In this case, you can buy chocolates priced at 1 and 3, leaving you with 4 - (1 + 3) = 0 units of money.
• Input: Input: prices = [5, 8, 7], money = 10
• Explanation: In this case, even though you have 10 units of money, the two cheapest chocolates cost 5 and 7, summing to 12, which exceeds the money you have. Therefore, you can't buy two chocolates, and the result is 10.
Approach: To solve the problem efficiently, we need to identify the two cheapest chocolates and compare their sum to the available money. If the sum is affordable, subtract it from the available money; otherwise, return the original money.
Observations:
• The problem is about selecting the two lowest prices and checking if they fit within the budget.
• Sorting the prices array to easily access the two smallest prices can be a quick solution.
Steps:
• Step 1: Sort the prices array in ascending order.
• Step 2: Check if the sum of the two smallest prices is less than or equal to the available money.
• Step 3: If the sum fits within the budget, return the remaining money after buying the two chocolates.
• Step 4: If the sum exceeds the available money, return the original amount of money.
Empty Inputs:
• There are always at least two prices, so no need to handle empty inputs.
Large Inputs:
• The solution should handle the upper limit of 50 chocolate prices efficiently.
Special Values:
• If the two cheapest chocolates' prices exceed the available money, return the original money.
Constraints:
• The solution should work efficiently within the constraints of prices and money.
int buyChoco(vector<int>& p, int m) {
int l1 = 200, l2 = 200;
for(int i = 0; i < p.size(); i++) {
if(p[i] < l1) {
l2 = l1;
l1 = p[i];
} else if(p[i] < l2) {
l2 = p[i];
}
}
if(l1 + l2 <= m) return m - l1 - l2;
return m;
}
1 : Function Declaration
int buyChoco(vector<int>& p, int m) {
This line declares the function `buyChoco`, which takes a reference to a vector `p` (prices of chocolates) and an integer `m` (the money available). The function returns an integer representing the remaining money after purchasing the two cheapest chocolates.
2 : Variable Initialization
int l1 = 200, l2 = 200;
Two variables `l1` and `l2` are initialized to 200. These represent the two lowest prices found in the list `p`.
3 : For Loop
for(int i = 0; i < p.size(); i++) {
A for loop is started to iterate through all the prices in the vector `p`.
4 : If Condition - First Lowest Price
if(p[i] < l1) {
This condition checks if the current price `p[i]` is smaller than the current lowest price `l1`.
5 : Update Variables
l2 = l1;
If the current price is the smallest, the second lowest price `l2` is updated to the previous lowest price `l1`.
6 : Update Variables
l1 = p[i];
The lowest price `l1` is updated to the current price `p[i]`.
7 : Else If Condition - Second Lowest Price
} else if(p[i] < l2) {
This condition checks if the current price is smaller than `l2` but greater than or equal to `l1`.
8 : Update Second Lowest Price
l2 = p[i];
If the current price is between `l1` and `l2`, the second lowest price `l2` is updated to the current price `p[i]`.
9 : If Condition - Purchase Feasibility
if(l1 + l2 <= m) return m - l1 - l2;
This condition checks if the sum of the two lowest prices is less than or equal to the available money `m`. If so, it returns the remaining money after purchasing the two chocolates.
10 : Return Statement - No Purchase
return m;
If the two lowest prices are more than the available money, the function returns the original amount of money `m` (no purchase is made).
Best Case: O(n log n)
Average Case: O(n log n)
Worst Case: O(n log n)
Description: The time complexity is O(n log n) due to the sorting step, where n is the number of chocolate prices.
Best Case: O(1)
Worst Case: O(n)
Description: The space complexity is O(n) for storing the prices array.
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