Pig Latin is a constructed language game in which words in English are altered, usually by adding a fabricated suffix or by moving the initial consonant or consonant cluster of a word to the end of the word and adding a vocalic syllable to create such a suffix. For example, Wikipedia would become ikipediaway (taking the 'W' and 'ay' to create a suffix). The objective is to conceal the words from others not familiar with the rules.” - Wikipedia

In this challenge, we will implement a JavaScript function that translates any English word supplied to Pig Latin.

Pretty cool! Yeah?

You should already have your development environment setup for this course. Open the cloned repository and inside the Beginner folder, navigate to the pigLatin folder. Our work for this challenge will be done in here. Make sure to follow along in the index-START.js file.

Translate the provided string to Pig Latin by following the rules below:

  • For words that begin with consonant sounds, the consonant before the initial vowel should be moved to the end of the word sequence and "ay" affixed. E.g
    • "pig" = "igpay"
  • For words that begin with consonant clusters, the clusters should be moved to the end of the word sequence and "ay" affixed. E.g
    • "glove" = "oveglay"
  • For words that begin with vowel sounds, simply add "way" to the end of the word. E.g
    • "explain" = "explainway”

This is a relatively difficult challenge, however walking through the rules will help us map things out better.

First, we need to test the first letter of the word to see if it is a consonant or vowel. If it is a vowel, then our job is easy. We simply add "``way``" to the end of the word.

However, if it is a consonant, it gets a little tricky. We need to determine if it stands alone or with another consonant or other consonants to form a cluster. This requires that we continue to test subsequent characters until we arrive at the first vowel. When we find the first vowel, we then move the identified consonant(s) to the end of the word and affix "``ay``".

Perhaps that wasn't very clear. Let's write some code.

We will consider two(2) ways to implement this function in JavaScript. They are:

  • An imperative approach
  • A declarative approach

Before going ahead to implement both solutions, it'd be helpful to gain a better understanding of the two terms used above.

Very often, we find these terms thrown around like they are very simple concepts everyone should know. However, the difference is usually not much obvious to most.

Simply put, an imperative style of programming is one which specifies how things get done. Although this might sound like what you do each time you write code, there's a difference to it. Imagine you were to add an array of numbers and return the sum, there are different ways you could approach the problem. One way could be writing a forloop that'd go over each element in the array and cumulatively add every element to an accumulator until the final sum is reached. That is imperative. You are specifying how things get done.

On the other hand, a declarative approach would abstract this process, allowing you to specify what should be done rather than how. Thus, you may use the .reduce() method on the array to reduce every element to a final value by returning the sum within the call back.

The point is, in imperative programming, you specify how things get done(step-by-step). However, in declarative programming, you simply state(declare) what should be done and aren't concerned with how it gets done.

Okay! We got that cleared up. Let's head back to solving this challenge.

This is a somewhat manual approach where we specify how things are done until we arrive at the final result. We make use of the for…of loop in iterating through the string in order to determine where the first vowel in the word occurs. See the solution below:

function pigLatin(str) {
  // Convert string to lowercase
  str = str.toLowerCase()
  // Initialize array of vowels
  const vowels = ["a", "e", "i", "o", "u"];
  // Initialize vowel index to 0
  let vowelIndex = 0;

  if (vowels.includes(str[0])) {
    // If first letter is a vowel
    return str + "way";
  } else {
    // If the first letter isn't a vowel i.e is a consonant
    for (let char of str) {
      // Loop through until the first vowel is found
      if (vowels.includes(char)) {
        // Store the index at which the first vowel exists
        vowelIndex = str.indexOf(char);
    // Compose final string
    return str.slice(vowelIndex) + str.slice(0, vowelIndex) + "ay";

In the code snippet above, we start by converting the received string str to lowercase. This is to prevent any casing related errors during comparison(“a” does not equal “A”).

Next, we initialize two variables:

  • vowels - containing the five English vowels
  • vowelIndex - for storing the index at which the first vowel in the word is found. It is initialized to 0.

We use an if…else statement to check if the first letter of the word can be found within our vowels array by calling the .includes() method on the array while passing it the first letter of the string str[0]. If it is found, this returns true, which implies that the first letter is a vowel. Hence, we simply add "``way``" to the end of the string and return the result as the Pig Latin equivalent.

If the statement evaluates to false, it signifies that the starting character is a consonant. Hence, we use a for…of loop to iterate through the string to identify the position of the first vowel. When we locate the first vowel, we use the .indexOf() method to retrieve it's position in the string and store it into the variable vowelIndex. After this step we terminate the loop using the break statement.

At the last line, we use the .slice() method to manipulate the string to generate the Pig Latin equivalent.

The .slice() method is used to extract a portion of an array starting from the specified beginning to end without modifying the original array. It returns a new array containing all elements from the starting index till the end index without including the item at the end index. When no end index is specified it assumes the end of the string as the end index.

  • str.slice(vowelIndex) extracts a portion of the string starting from the first vowel(i.e the vowelIndex up till the end.
  • str.slice(0, vowelIndex) extracts the portion of the string starting at index 0(i.e the beginning) up until the first vowel at position vowelIndex. The extracted portion here is the consonant or consonant cluster as the case may be. Note that the .slice() method returns the portion from the starting index to but not including the end index.

We combine these extracted portions and affix "``ay``" to the end to form the final result which is returned from the function as such.

Now let's get a lil lazy. 😈

In this approach, we implement a very concise solution to this challenge by combining the .replace() method and regular expressions to transform the received string into its Pig Latin equivalent.

The replace() method returns a new string with some or all matches of a pattern replaced by a specified replacement.

function pigLatin(str) {
    return str
    .replace(/^([aeiouy])(._)/, '$1$2way')
    .replace(/^(_[_^aeiouy]+)(._)/, '$2$1ay')

In this challenge, we cannot afford to dive deeply into the mechanics of regular expressions as that is a topic on it own. However, we will make sure to sufficiently explain what is going on in the solution above.

Our solution comprises mainly of two parts as analyzed below:

  • .replace(/^([aeiouy])(.*``*)/, '$1$2way')* - This statement specifies a replacement to be carried out if the word begins with a vowel. This is specified in the first bracket within the *.replace()* method call i.e *([aeiou])*. The second bracket *(.*)* refers to every other character after the vowel. Thus, the expression specifies a pattern for words beginning with a vowel and followed by anything else. When this case is matched, a new string in the format of '``$1$2way``' is generated and used to replace the original srtring. $1 here refers to the first bracket and $2, the second bracket. This means that we simply take the word as it was and affix "``way``" to the end.
  • .replace(/^([^aeiouy]+)(.*)/, '$2$1ay') - This statement specifies that if the word does not start with a vowel i.e ^([aeiouy]+), and is followed by anything else (``*.*)*, it should be replaced with a string formatted in the order '$2$1ay``'. The plus sign in ^([aeiouy]+) caters for a situation where there is a consonant cluster. Thus it represents every non-vowel character at the start of the word. '$2$1ay' generates the new string in the order of remaining characters + consonant cluster + '``ay``'. This gives the Pig Latin equivalent.

Note that we chain both .replace() methods in succession such that both cases are tested and only the one that matches will be evaluated further.

For a deeper understanding of regular expressions, you may use the resource cited at the end of this challenge.

To run the tests for this challenge, simply open your terminal and execute the command below:

npm run test pigLatin

Awesome, we passed all tests(as always 😉 ).

To run the performance test for this challenge, follow this link to JSPerf. The screenshot below, reveals the result of our comparison.

The result of our comparison reveals that the fastest solution is the Imperative approach and the declarative approach is approximately 88% slower.

Pig Latin although not a formal language is mostly used as an obfuscation technique in games. This means that it is used to pass messages across in a coded form, thus reducing the readability and preventing it from being easily interpreted.

It may also be encountered as an interview challenge testing ones string manipulation capabilities.

In the course of solving this challenge, we have successfully implemented a function that returns the Pig Latin equivalent of any English word received.

Our performance comparison reveals that the imperative approach is the optimal solution. This in no way implies that the imperative style of programming is generally more optimized than the declarative.

Interestingly, this challenge also reveals that fewer lines of code doesn't directly translate to better code.

For a deeper understanding of the concepts and techniques highlighted above, you may use the following links:

See you in the next one!✌🏿

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