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 balance and how much more money you need to buy one additional box. This problem is significant in scenarios involving budgeting and financial planning.
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 the following 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 amount 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 desired result efficiently.
Here is a step-by-step breakdown of the algorithm:
boxes = balance / price
.balanceNeeded = (boxes + 1) * price
.additionalAmount = balanceNeeded - balance
.public class BuyCandy {
public static int additionalAmountNeeded(int price, int balance) {
// Calculate the number of boxes that can be bought with the current balance
int boxes = balance / price;
// Calculate the total cost needed to buy one more box
int balanceNeeded = (boxes + 1) * price;
// Calculate the additional amount needed to buy one more box
int additionalAmount = balanceNeeded - balance;
return additionalAmount;
}
public static void main(String[] args) {
int price = 4;
int balance = 9;
System.out.println(additionalAmountNeeded(price, balance)); // Output: 3
}
}
The time complexity of this solution is O(1) because the operations involved (division, multiplication, and subtraction) are constant time operations.
The space complexity is also O(1) as we are using a fixed amount of extra space for variables.
Consider the following edge cases:
balance = 0
, the output should be the price of one box.balance
is an exact multiple of price
, the additional amount needed should be the price of one box.Example:
Input: price = 4, balance = 0 Output: 4 Input: price = 4, balance = 8 Output: 4
To test the solution comprehensively, consider a variety of test cases:
When approaching such problems, consider breaking down the problem into smaller, manageable steps. Use integer division and modular arithmetic to simplify calculations. Practice similar problems to improve your problem-solving skills.
In this blog post, we discussed how to solve the problem of determining the number of candy boxes you can buy and the additional amount needed for one more box. We explored a step-by-step approach, provided a Java implementation, and analyzed the complexity. Understanding and solving such problems is crucial for developing strong problem-solving skills.
For further reading and practice, consider the following resources: