Hello, checkiomates🐱‍👤!

This week, we elevate your code quality by exploring secure coding in JavaScript, dive into advanced type systems beyond TypeScript, and learn how to catch UI bugs early using Playwright for visual regression testing. Our mission challenges your logic to analyze an integer sequence and count the number of times the sorting direction changes.

💡TIP

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🏁MISSION

Changing direction by freeman_lex -

You are given a sequence of integers. Your task in this mission is to find, how many times the sorting direction was changed in the given sequence. If the elements are equal - the previous sorting direction remains the same, if the sequence starts from the same elements - look for the next different to determine the sorting direction.

assert.strictEqual(changingDirection([1, 2, 3, 4, 5]), 0);
assert.strictEqual(changingDirection([1, 2, 3, 2, 1]), 1);
assert.strictEqual(changingDirection([1, 2, 2, 1, 2, 2]), 2);

📖ARTICLES

Secure coding in JavaScript -

JavaScript is the front-end of the entire internet. Whether you transpile TypeScript down into JavaScript, create fast little node.js scripts, or build a beautiful-but-dumb front end that calls a much more interesting collection of APIs, it’s literally everywhere. Because JavaScript is so prolific, it’s a prime target for attackers. In this article we will cover ten tips for writing more secure JavaScript.

Visual Types -

A handy visually-oriented guide/tour of what types are and how they work, with a focus on TypeScript.

How We Catch UI Bugs Early with Visual Regression Testing -

Visual regression testing is a powerful technique to ensure our web application looks as expected, even as code changes over time. In this article, we walk through what a visual regression testing tool is, why it’s important, and how it's implemented in CI/CD pipeline using Playwright, GitHub Actions, and Git LFS

👩‍💻CODE SHOT

What do you think the following code does?

function checkio(values: number[]): boolean {
    
    let cur = NaN;
    for (let i of values){
        if (cur && i <= cur) return false;
        cur = i;
    }
    return true;
}

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