freeCodeCamp Challenge Guide: Pig Latin

freeCodeCamp Challenge Guide: Pig Latin
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Pig Latin


Problem Explanation

You need to create a program that will translate from English to Pig Latin. Pig Latin takes the first consonant (or consonant cluster) of an English word, moves it to the end of the word and suffixes an “ay”. If a word begins with a vowel you just add “way” to the end. It might not be obvious but you need to remove all the consonants up to the first vowel in case the word does not start with a vowel.

Relevant Links


Hints

Hint 1

You will probably want to use regular expressions. This will allow you to convert the words easily.

Hint 2

If the first character is a vowel, then take that whole word and add ‘way’ at the end. Otherwise comes the tricky part, take the consonant(s) before the first vowel and move it to the end and add ‘ay’. This might be confusing but, it is not just the first consonant but all of them before the first vowel.

Hint 3

You will need to use everything you know about string manipulation to get the last part right. However, it can be done with substr alone.


Solutions

Solution 1 (Click to Show/Hide)
function translatePigLatin(str) {
  let consonantRegex = /^[^aeiou]+/;
  let myConsonants = str.match(consonantRegex);
  return myConsonants !== null
    ? str
        .replace(consonantRegex, "")
        .concat(myConsonants)
        .concat("ay")
    : str.concat("way");
}

translatePigLatin("consonant");

Code Explanation

  • start at beginning and get longest match of everything not a vowel (consonants)

  • if regex pattern found, it saves the match; else, it returns null

  • if regex pattern found (starts with consonants), it deletes match, adds the match to the end, and adds “ay” to the end

  • if regex pattern not found (starts with vowels), it just adds “way” to the ending

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Solution 2 (Click to Show/Hide)
function translatePigLatin(str) {
  // Create variables to be used
  var pigLatin = "";
  var regex = /[aeiou]/gi;

  // Check if the first character is a vowel
  if (str[0].match(regex)) {
    pigLatin = str + "way";
  } else if (str.match(regex) === null) {
    // Check if the string contains only consonants
    pigLatin = str + "ay";
  } else {
    // Find how many consonants before the first vowel.
    var vowelIndice = str.indexOf(str.match(regex)[0]);

    // Take the string from the first vowel to the last char
    // then add the consonants that were previously omitted and add the ending.
    pigLatin = str.substr(vowelIndice) + str.substr(0, vowelIndice) + "ay";
  }

  return pigLatin;
}

// test here
translatePigLatin("consonant");

Code Explanation

  • Make an empty string to hold your Pig Latin word.
  • Assign your appropriate regular expression to a variable.
  • If the first character is a vowel, just add way to end of string and return it.
  • If the first character is not a vowel:
    • Find number of consonants before first vowel with help of indexOf(), match() and regex.
    • Start Pig Latin string with first vowel till the end.
    • Add letters before first vowel to end of string.
    • substr() is used for string manipulation here.
    • Add ay to end of string and return it.

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Solution 3 (Click to Show/Hide)
function translatePigLatin(str) {
  if (str.match(/^[aeiou]/)) return str + "way";

  const consonantCluster = str.match(/^[^aeiou]+/)[0];
  return str.substring(consonantCluster.length) + consonantCluster + "ay";
}

// test here
translatePigLatin("consonant");

Code Explanation

  • First, check to see if the string begins with a vowel.
    • The regex looks at the beginning of the string ^ for one of the specified characters [aeiou]
    • If it does, you only need to return the original string with “way” appended on the end.
  • If the string does not start with a vowel, we want to build a string which contains every consonant before the first vowel in the provided string.
    • To do this, look at the beginning of a string ^ for one or more characters + NOT specified [^aeiou].
    • If there is a match (and in this case, there always will be), match() returns an Array with the matched string as the first element, which is all we want. Grab it with [0].
  • Now, we can start building our Pig Latin string to return. This can be built in three parts:
    • The first part contains all of the characters in the original string, starting from the first vowel. We can easily get these characters by creating a substring of the original string, with its starting index being the first vowel.
    • The second part contains the consonant string we just built. (If you add the second and first parts of this string together, you will get the original string.)
    • The final part contains “ay”.

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Solution 4 (Click to Show/Hide)
function translatePigLatin(str) {
  return str
    .replace(/^[aeiou]\w*/, "$&way")
    .replace(/(^[^aeiou]+)(\w*)/, "$2$1ay");
}

// test here
translatePigLatin("consonant");

Code Explanation

  • Use replace() on the string, using a regular expression to check if the first letter is a consonant and adding way at the end in this case. If the first letter is a consonant nothing will happen at this point.
  • Use replace() again to check for consonants at the beginning of the word and to move it or them to the end of the word and add ay at the end.

Relevant Links

6 Likes

Thanks for sharing information!

BTW My code is following:

function translatePigLatin(str) {
  var newString = str + "way";
  
  if (/[qwrtypsdfghjklzxcvbnm]{1,}/i[Symbol.match](str[0])) {
  var firstConsonants = /[qwrtypsdfghjklzxcvbnm]{1,}/i[Symbol.match](str);
  var length = firstConsonants[0].length;
  var newStr = str.substr(length) + firstConsonants + "ay";
  return newStr; 
    
  }
  
  return newString;

}

translatePigLatin("glove");
2 Likes

For what it’s worth, my solution is similar to the intermediate one, but I used String.prototype.search() to find the index of the first vowel:

function translatePigLatin(str) {
  var firstVowel = str.search(/[aeiou]/);
  
  return firstVowel === 0 ? str + 'way' : str.substr(firstVowel) + str.substr(0, firstVowel) + 'ay';
}
15 Likes

In the advanced code solution :rotating_light: , and also in @vkg’s solution, the regEx should be /[aeiou]/i because without the i modifier the test will be case sensitive and strings starting with a capital vowel will be translate as consonant starting (e.g. translatePigLatin(“Airway”); will return “irwayAay” instead of “Airwayway”).

2 Likes

Didn’t see any solution with just string.prototype.replace(); so I thought I’d add mine.

function translatePigLatin(str) {
  if (["a","e","i","o","u"].indexOf(str.charAt(0)) != -1){
    return str += "way";
  }
  return str.replace( /([^aeiou]*)([aeiou]\w*)/ , "$2$1ay");
}

translatePigLatin("consonant");
7 Likes

As if this writing, the description on freecodecamp for this exercise states:

“Input strings are guaranteed to be English words in all lowercase.”

1 Like

This should be the recommended advanced solution. Not only is it concise and easy to read, but it also works. Whereas the other solutions presented fail on consonant only words, such as ‘my’ or ‘TV’.

my version of pig latin. i did not use regular express. however, the regular express solution are pretty sweet!

function translatePigLatin(str) {
var vowels= [‘a’,‘e’,‘i’,‘o’,‘u’], index=0;

 if(vowels.indexOf(str[index])!== -1){  //initial edge case
     return str+"way";
 }
 for(index=1;index<str.length;index++){
     if(vowels.indexOf(str[index]) !==-1){//first vowel found
         break;
    }
 }

return str.slice(index,str.length)+str.slice(0, index)+“ay”;
}

Some good solutions on here. Mine’s a little different, so I’ll post it too. I used String.prototype.replace(), with a function rather than direct string replacement:

function translatePigLatin(str) {
  return str.replace(/(^[^aeiou]*)(\w*)/, function(match, p1, p2){
    return p2 + (p1 ? p1 : 'w') + 'ay';
  })
}

edit to add explanation:
Regarding the arguments passed to the inline function:

  • match in this case only serves as a placeholder since it is always the first argument passed by replace( ) but is not actually used in the function (with this regex match is always the whole word (given the input string is stipulated to be an English word in lowercase).
  • p1 is the first parenthesized submatch string ^[^aeiou]* , i.e. zero or more non-vowels at the beginning of the word. Ex. for ‘glove’ p1 = ‘gl’ and for ‘algorithm’ p1 = “”.
  • p2 will be all the characters in the word that were not captured by p1. Ex. ‘glove’ p2 = ‘ove’ and for ‘algorithm’ p2 = ‘algorithm’.
4 Likes

The Basic solution should be extended by an if construct that first checks if there are any matches before it tries to assign them to vowelIndice. Its a pretty big oversight imho.

Otherwise the solution is perfectly fine.

Hello, all your solutions looks very clever, my solution is probably a bit basic, because I use only for loops and if’s to check for the vowels. And I haven’t heard of regular expressions prior to reading the solution. So what do you think of my solution?

      function translatePigLatin(str) {

      var vowels = ["a","e","i","o","u"];
      var strArr = str.split("");
      var index;

      // Look for the vowel index in a string
      for(var k = 0; k < strArr.length; k++){
        for(var i = 0; i < vowels.length; i++){
          if(strArr[k] === vowels[i]){
            if(index === undefined){
              index = strArr.indexOf(strArr[k]);
            }
          }
        }
      }

      // if vowel is not the first letter use pig latin method
      if(index !== 0){
        var sliced = str.slice(0, index);
        str = str.slice(index) + sliced + "ay";
      }else {
        str = str + "way";
      }

      return str;
    }

    translatePigLatin("consonant");
2 Likes

Another solution. Using indexOf, charAt and substr functions.

function translatePigLatin(str) {
var vow = ['a','e','i','o','u'];
if(vow.indexOf(str.charAt(0)) !== -1){
  str += "w";
}
while(vow.indexOf(str.charAt(0)) == -1){
  str = str.substr(1) + str.charAt(0);
}
return str += "ay";
}
//test here
translatePigLatin("california");

How should I understand this return statement? Does it return

['a','i','u','e','o'].indexOf(str.charAt(obj)) == -1 ? check(obj + 1) : obj;

or does it just return
check(obj + 1) or obj depending on the result of ['a','i','u','e','o'].indexOf(str.charAt(obj)) == -1 ?

My solution is ultra-basic so I appreciate these threads so I can take notes for improving my approach for later exercises.

That said, I have to point out that none of the solutions I’ve read in this thread so far appear to address the ‘y’ character; i.e., the word ‘dynamic’ in pig latin should be ‘ynamicday’ but unless I’m missing something, every solution here would return ‘amicdynay’. Again, unless I’m missing something.

In any case, here’s my red-headed stepchild:


function translatePigLatin(str) {
  var vowels = ['a', 'e', 'i', 'o', 'u', 'y'];
  var str1 = "";
  
  if(vowels.indexOf(str.charAt(0)) !== -1 && str.charAt(0) !== "y") {
    str1 += str + "way";
  } else if(str.charAt(0) === "y") {
    str1 += str.slice(1) + "yay";
  } else {
    for(var i = 0; i < str.length; i++) {
      if(vowels.indexOf(str.charAt(i)) !== -1) {
        return str1 += str.slice(i) + str.slice(0, i) + "ay";
      }
    }
  }
  str = str1;
  return str;
}

Ladies and Gentlemen, as a long time reader but a first time poster on this forum I present to you the world most inelegant code that solve the pigLatin algorithm.

function translatePigLatin(str) {
var chr = "";
for (var i = 0; i < str.length; i++) {
    chr = str.charAt(i);
    if (chr == "a" || chr == "e" || chr == "i" || chr == "o" || chr == "u") {
        break;
    }
}
//console.log(str.indexOf(chr)); //this gives the position of the first vowel in the word.
if (chr === str[0]) {
    str = str + "way";
    console.log(str); //use return on FCC
} else {
    str = str.split("");
    //the consonant or consonant clusters are stored in removed
    var removed = str.slice(0, str.indexOf(chr)).join("");
    //the rest of the word starting with the vowel is stored in vowelInitial
    var vowelInitial = str.slice(str.indexOf(chr)).join("");
    var allTogether = vowelInitial + removed + "ay";

    console.log(allTogether); //use return on FCC
}

}

Solution using just match and replace with regex

function translatePigLatin(str) {
  if(str.match(/^[aeiou]/)){  //starts with vowel
    return str.replace(/(.+)/,"$1way");
  }else if(str.match(/[aeiou]/g)){ //starts with consonant or consonant cluster
    return str.replace(/(^[^aeiou]+)(.+)/g,"$2$1ay");
  }else{ //all consonants 
    return str + "ay";
  }
}

My solution

function translatePigLatin(str) {
  var strArr = str.split('');
  var frstLet = strArr[0];
  var pigArr = [];
  var i = 1;
  
  function checkVowel(x) {
    return (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u');
  }

  if (checkVowel(frstLet)) {
    strArr.push('way');
    str = strArr.join('');
  } else {
    pigArr.push(frstLet);
    while (!checkVowel(strArr[i])) {
      pigArr.push(strArr[i]);
      i++;
    }
    pigArr.push('a', 'y');
    pigArr = pigArr.join('');
    str = strArr.slice(i).join('') + pigArr;
  }
  
  return console.log(str);
}

My solution:
function translatePigLatin(str) {
var vowel = [“a”,“o”,“i”,“e”,“y”,“u”];
var cluster = str.charAt(0) + str.charAt(1);
if (vowel.indexOf(str[0]) !== -1) {
return str + ‘way’;
}
else if (vowel.indexOf(str[0]) == -1 && vowel.indexOf(str[1]) == -1) {
return str.substring(2) + cluster + ‘ay’;
}
else{
return str.slice(1) + str[0] + ‘ay’;
}
}