A box of candy costs price dollars. You have balance dollars. Compute the number of boxes of candy you can buy and return how many more dollars you need to buy one more box of candy.
Example:
Input: price = 4, balance = 9 Output: 3 Explanation: You can buy two boxes of candy that costs 4 * 2 = 8 dollars. If you would have 12 dollars you could buy 3 boxes (one more), so you would need 3 more dollars.
The core challenge of this problem is to determine how many boxes of candy you can buy with a given amount of money and how much more money you would need to buy one additional box. This problem is significant in scenarios where budgeting and incremental purchasing decisions are involved.
Potential pitfalls include miscalculating the number of boxes you can buy or the additional amount needed due to incorrect arithmetic operations.
To solve this problem, we can break it down into a few simple steps:
Let's start with a naive approach and then optimize it.
The naive approach involves directly calculating the number of boxes and the additional money needed. This approach is straightforward but not necessarily the most efficient in terms of readability and maintainability.
The optimized approach involves using integer division and simple arithmetic operations to achieve the same result in a more concise manner.
Here is a step-by-step breakdown of the optimized algorithm:
boxes = balance / price.balanceNeeded = (boxes + 1) * price.additionalMoney = balanceNeeded - balance.
#include <iostream>
using namespace std;
int additionalMoneyNeeded(int price, int balance) {
// Calculate the number of boxes that can be bought
int boxes = balance / price;
// Calculate the total cost for one more box
int balanceNeeded = (boxes + 1) * price;
// Calculate the additional money needed
int additionalMoney = balanceNeeded - balance;
return additionalMoney;
}
int main() {
int price = 4;
int balance = 9;
cout << additionalMoneyNeeded(price, balance) << endl; // Output: 3
return 0;
}
The time complexity of this approach is O(1) because it involves a constant number of arithmetic operations. The space complexity is also O(1) as no additional space is required beyond a few integer variables.
Consider the following edge cases:
balance is less than price, the function should return price - balance.balance is exactly a multiple of price, the function should return price.These edge cases are handled naturally by the algorithm due to the arithmetic operations used.
To test the solution comprehensively, consider the following test cases:
Test Case 1: Input: price = 4, balance = 9 Output: 3 Test Case 2: Input: price = 5, balance = 20 Output: 5 Test Case 3: Input: price = 7, balance = 6 Output: 1 Test Case 4: Input: price = 3, balance = 15 Output: 3
These test cases cover a range of scenarios from having just enough balance to needing a small or large additional amount.
When approaching such problems, it's essential to break down the problem into smaller, manageable parts. Focus on understanding the requirements and constraints first. Practice similar problems to improve your problem-solving skills and familiarize yourself with common patterns and techniques.
In this blog post, we discussed how to solve the problem of determining how many boxes of candy you can buy with a given balance and how much more money you need to buy one more box. We explored a step-by-step approach, provided a detailed algorithm, and implemented the solution in C++. Understanding and solving such problems is crucial for developing strong problem-solving skills in programming.
For further reading and practice, consider the following resources: