Section 1.1: The Problem Statement

Scenario: Purchasing a Car

A man owns an older vehicle valued at $2000 and has his sights set on a secondhand car priced at $8000. He plans to retain his current car until he has saved enough to purchase the new one. He estimates that he can save $1000 each month; however, both the value of his old car and the price of the new car depreciate by 1.5% monthly. Additionally, this depreciation rate increases by 0.5% every two months, complicating his savings calculations.

Can you assist him in figuring out how long it will take to save enough money to buy the desired car and how much he will have left afterward?

Function Parameters and Return:

• start_price_old: (Positive float/int) The price of the old car.
• start_price_new: (Positive float/int) The price of the new car.
• saving_per_month: (Positive float/int) Monthly savings amount.
• percent_loss_by_month: (Positive float/int) The monthly depreciation percentage.

For instance, calling nbMonths(2000, 8000, 1000, 1.5) should yield [6, 766], indicating he can purchase the new car in 6 months with $766 remaining.

Example Breakdown:

• End of Month 1: Percent loss: 1.5% | Amount available: -4910.0
• End of Month 2: Percent loss: 2.0% | Amount available: -3791.7999
• End of Month 3: Percent loss: 2.0% | Amount available: -2675.964
• End of Month 4: Percent loss: 2.5% | Amount available: -1534.06489
• End of Month 5: Percent loss: 2.5% | Amount available: -395.71327
• End of Month 6: Percent loss: 3.0% | Amount available: 766.158120825

Thus, the final return would be [6, 766] — 6 months to save up, and approximately $766 remaining after the purchase.

Section 1.2: Implementation of the Solution

My Approach

Here’s one way to structure the solution:

export function nbMonths(params) {

while (getMoneyLeft(params) < 0) {

params = { ...updateValues(params) };

}

return [params.month, Math.round(getMoneyLeft(params))];

}

Alternatively, a recursive approach could be utilized:

export function nbMonths(params) {

const moneyLeft = getMoneyLeft(params);

if (moneyLeft >= 0) {

return [params.month, Math.round(getMoneyLeft(params))];

} else {

return nbMonths(updateValues(params));

}

}

To keep the code organized, I opted to separate the problem into smaller functions, enabling clearer logic flow.

Section 1.3: Breaking Down the Functions

To improve readability and manageability, I defined a parameter object:

type Param = {

priceOld: number;

priceNew: number;

saving: number;

percentLoss: number;

month: number;

};

Next, I implemented a function to calculate the updated prices of the two cars:

const value = (price, delta) => price * (1 - delta / 100);

Subsequent functions update the prices and the loss percentage based on the current month:

const updatePriceNew = (p) => value(p.priceNew, p.percentLoss);

const updatePriceOld = (p) => value(p.priceOld, p.percentLoss);

const updatePercentLoss = (p) => p.percentLoss + (1 - (p.month % 2)) * 0.5;

const updateMonth = (p) => p.month + 1;

Finally, I created a function to calculate how much money remains after purchasing the car:

const getMoneyLeft = (p) => p.month * p.saving + p.priceOld - p.priceNew;

The final assembly of these components leads us back to the complete solution.