Weekly Coding Challenge: Buying a car – written in JavaScript and Python

Posted by

2 min read ~ Hello readers! It’s really important for me to continue my practice and learning. Lately, I have been completing a coding challenge a day with codewars. This helps me keep my skills sharp and see how other devs all over the planet would solve these problems. Being able to see other solutions expands my mind for solving future problems in a clever and efficient way. I’ve decided that I am going to post my solutions here written in JavaScript and Python for all of my readers to learn! Please let me know if you have any questions.

Buying a Car

Let us begin with an example:

A man has a rather old car being worth $2000. He saw a secondhand car being worth $8000. He wants to keep his old car until he can buy the secondhand one.

He thinks he can save $1000 each month but the prices of his old car and of the new one decrease of 1.5 percent per month. Furthermore this percent of loss increases of 0.5 percent at the end of every two months. Our man finds it difficult to make all these calculations.

Can you help him?

How many months will it take him to save up enough money to buy the car he wants, and how much money will he have left over?

Parameters and return of function:

parameter (positive int or float, guaranteed) startPriceOld (Old car price)
parameter (positive int or float, guaranteed) startPriceNew (New car price)
parameter (positive int or float, guaranteed) savingperMonth 
parameter (positive float or int, guaranteed) percentLossByMonth

nbMonths(2000, 8000, 1000, 1.5) should return [6, 766] or (6, 766)

Detail of the above example:

end month 1: percentLoss 1.5 available -4910.0
end month 2: percentLoss 2.0 available -3791.7999...
end month 3: percentLoss 2.0 available -2675.964
end month 4: percentLoss 2.5 available -1534.06489...
end month 5: percentLoss 2.5 available -395.71327...
end month 6: percentLoss 3.0 available 766.158120825...
return [6, 766] or (6, 766)

where 6 is the number of months at the end of which he can buy the new car and 766 is the nearest integer to 766.158... (rounding 766.158 gives 766).


Selling, buying and saving are normally done at end of month. Calculations are processed at the end of each considered month but if, by chance from the start, the value of the old car is bigger than the value of the new one or equal there is no saving to be made, no need to wait so he can at the beginning of the month buy the new car:

nbMonths(12000, 8000, 1000, 1.5) should return [0, 4000]
nbMonths(8000, 8000, 1000, 1.5) should return [0, 0]

JavaScript solution:

function nbMonths(startPriceOld, startPriceNew, savingperMonth, percentLossByMonth){

    // Check to see you even need to save first! If not, return out
    if(startPriceOld >= startPriceNew) {return [0, Math.round(startPriceOld - startPriceNew)];}
    // Otherwise, declare the months and the actual total 
    var months = 0, total = startPriceOld;
    // while loop to run while the total is less than the new price of the car
    while(total < startPriceNew) {
        // increment the savings per month onto the total
        total += savingperMonth;
        // decrease the total by the percentage loss per month
        total -= startPriceOld * percentLossByMonth / 100;

        // manipulate the start price new and old to reflect the percent loss by the current month
        startPriceNew -= startPriceNew * percentLossByMonth / 100;
        startPriceOld -= startPriceOld * percentLossByMonth / 100;
        // Increment the months
        months ++;

        // Use mod to determine even or odd month so that we know if we should increase the percentage loss by month by .5
        months % 2 !== 0 ? percentLossByMonth += 0.5 : percentLossByMonth;
  return [months, Math.round(total - startPriceNew)];

Python solution:

def nbMonths(startPriceOld, startPriceNew, savingperMonth, percentLossByMonth):
    if startPriceOld >= startPriceNew:
        return [0, round(startPriceOld - startPriceNew)] 
    months = 0
    total = startPriceOld
    while (total < startPriceNew):
        total += savingperMonth;
        total -= startPriceOld * (percentLossByMonth / 100)
        startPriceNew -= startPriceNew * (percentLossByMonth / 100)
        startPriceOld -= startPriceOld * (percentLossByMonth / 100)
        months = months + 1
        if months % 2 != 0:
            percentLossByMonth += 0.5
    return [months, round(total - startPriceNew)]

Disclaimer: *Now, as much as I am tempted to use ES8 & ES9 functions to reduce the number of lines in JavaScript, I am always a fan of readable code that others can understand. Therefore, with the JavaScript examples, I will try to write it without fancy functions as much as possible so it can actually be read and absorbed by my readers. All of my solutions are commented to explain my thinking. Of course, they could always be better. Please feel free to share how you would solve it!*

All credit to problem sets goes to codewars

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s