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 straightforward but requires careful handling of integer division and basic arithmetic operations.
Common applications of this problem include budgeting scenarios, resource allocation, and financial planning where you need to determine how much more of a resource is required to achieve a goal.
Potential pitfalls include incorrect handling of integer division and not properly calculating the additional amount needed for one more box.
To solve this problem, we can break it down into the following steps:
Let's start with a naive solution and then optimize it.
The naive solution involves directly performing the calculations as described. This approach is simple and works well for this problem since the operations involved are basic arithmetic.
The optimized solution follows the same steps but ensures that the calculations are done efficiently and correctly. Given the simplicity of the problem, the naive solution is already optimal.
Here is a step-by-step breakdown of the algorithm:
boxes = Math.floor(balance / price)
.balanceNeeded = (boxes + 1) * price
.additionalAmount = balanceNeeded - balance
.// Function to calculate the number of boxes and additional amount needed
function buyCandy(price, balance) {
// Calculate the number of boxes that can be bought
const boxes = Math.floor(balance / price);
// Calculate the total cost for one more box
const balanceNeeded = (boxes + 1) * price;
// Calculate the additional amount needed
const additionalAmount = balanceNeeded - balance;
// Return the additional amount needed
return additionalAmount;
}
// Example usage:
const price = 4;
const balance = 9;
console.log(buyCandy(price, balance)); // Output: 3
The time complexity of this solution is O(1) because it involves a constant number of arithmetic operations. The space complexity is also O(1) as it uses a fixed amount of additional space for variables.
Potential edge cases include:
Examples:
buyCandy(4, 3); // Output: 1
buyCandy(4, 8); // Output: 4
To test the solution comprehensively, consider the following test cases:
Example test cases:
console.log(buyCandy(4, 3)); // Output: 1
console.log(buyCandy(4, 8)); // Output: 4
console.log(buyCandy(4, 0)); // Output: 4
console.log(buyCandy(4, 12)); // Output: 4
When approaching such problems, consider the following tips:
To improve problem-solving skills, practice solving similar problems and study different algorithms and their applications.
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 provided a detailed explanation of the algorithm, code implementation, complexity analysis, and testing strategies. Understanding and solving such problems is crucial for developing strong problem-solving skills and improving algorithmic thinking.
For further reading and practice problems, consider the following resources: