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What’s your alternative solution? Challenge #12

codeguppy profile image Adrian ・2 min read

About this series

This is series of daily JavaScript coding challenges... for both beginners and advanced users.

Each day I’m gone present you a very simple coding challenge, together with the solution. The solution is intentionally written in a didactic way using classic JavaScript syntax in order to be accessible to coders of all levels.

Solutions are designed with increase level of complexity.

Today’s coding challenge

Create a function that receives an array of numbers and returns an array containing only the positive numbers

(scroll down for solution)

Code newbies

If you are a code newbie, try to work on the solution on your own. After you finish it, or if you need help, please consult the provided solution.

Advanced developers

Please provide alternative solutions in the comments below.

You can solve it using functional concepts or solve it using a different algorithm... or just solve it using the latest ES innovations.

By providing a new solution you can show code newbies different ways to solve the same problem.

Solution

// Solution for challenge12a

function getPositives(ar)
{
    var ar2 = [];

    for(var i = 0; i < ar.length; i++)
    {
        var el = ar[i];

        if (el >= 0)
        {
            ar2.push(el);
        }
    }

    return ar2;
}

var ar = [-5, 10, -3, 12, -9, 5, 90, 0, 1];
var ar2 = getPositives(ar);

println(ar2);

To quickly verify this solution, copy the code above in this coding editor and press "Run".

Note: The solution was originally designed for codeguppy.com environment, and therefore is making use of println. This is the almost equivalent of console.log in other environments. Please feel free to use your preferred coding playground / environment when implementing your solution.

Discussion

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aminnairi profile image
Amin

My take at the challenge using recursion.

/**
 * Takes only positives values from an array
 *
 * @params {number[]} numbers Numbers's array
 *
 * @throws {Error}      If the function is not called with one argument
 * @throws {TypeError}  If the first argument is not an array
 * @throws {TypeError}  If the first argument is not an array of numbers
 *
 * @return {number[]} Positives numbers
 *
 * @example
 * positivesOnly([-1, 1, -2, 2]);   // [1, 2]
 * positivesOnly([1, 2]);           // [1, 2]
 * positivesOnly([-1, -2]);         // []
 */
const positivesOnly = (numbers, ...extraParameters) => {
    if (extraParameters.length !== 0) {
        throw new Error("expected one argument");
    }

    if (!Array.isArray(numbers)) {
        throw new TypeError("expected first argument to be an array");
    }

    if (numbers.length === 0) {
        return [];
    }

    const [number, ...rest] = numbers;

    if (typeof number !== "number") {
        throw new TypeError("expected first argument to be an array of numbers");
    }

    if (number < 0) {
        return positivesOnly(rest);
    }

    return [number, ...positivesOnly(rest)];
};
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dannylee8 profile image
Danny Lee

Here's my alternative solution using es6 filter function and arrow functions.

It works by taking the 'ar' array and iterating through it (like a for loop would do) and instead of using the for loops "el[i]" syntax to show an individual element, for each iteration the item is just called "el". This could actually be named anything (el, element, num, numberToBeTested, etc). Then, the arrow function => preceeds what the filter() will return as an acceptable value. In this case, any iteration where el >= 0 will be stored and then once the entire loop is finished, the filter() will return an array of values that matched the condition. Hope that made some sense. This took me forever to understand so I get it if you're like '????' 😁

function getPositives(ar)
{
return ar.filter(el => el >= 0);
}

var ar = [-5, 10, -3, 12, -9, 5, 90, 0, 1];
var ar2 = getPositives(ar);

println(ar2);