Javascript: Coderbyte Challenge 4: Letter Changes: help analyze code: how this capitalizes vowels and when the else statement is executed

Question

I'm trying to analyze/reverse-engineer someone's Coderbyte code.The purpose of this challenge is to take a string parameter and to replace every letter of the string w/ the letter following it in the alphabet (a becomes b, z becomes a, etc). Then we are to capitalize every vowel in the new string (a, e, i, o, u). After finding someone's code, I was amazed at how he/she simplified it, but I had two specific questions:

1) what part of the code (see below) does the capitalizing??? I'm not seeing it so I must be missing it??? (watch it's probably going to be something completely obvious)

and

2) when would the else-statement come into play? the if-else is if (n>-1), so when would n <= -1? In the code, n is always going to be the index of a particular letter of the given string-parameter - it's index in oldAlph. That index would consist of 0 - 25. So I don't understand when it would be -1, or -2, etc...

Here is the code:

function LetterChanges(str) { 

  // code goes here
  var n = 0;
  var nstr = "";
  var oldAlph = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"];
  var newAlph = ["b","c","d","E","f","g","h","I","j","k","l","m","n","O","p","q","r","s","t","U","v","w","x","y","z","A"];
  for (i=0;i<str.length;i++) {
    n = oldAlph.indexOf(str.substr(i,1));
    if (n>-1) {
      nstr = nstr + newAlph[n];
    } else {
      nstr = nstr + str.substr(i,1);
    }
  }
   return nstr; 

}

// keep this function call here 
// to see how to enter arguments in JavaScript scroll down
LetterChanges(readline());

Answer 1

1: you are grabbing the capitalized version of the character and appending it to the new string here: nstr = nstr + newAlph[n];

2: it will be === -1 if the current character is not found within the oldAlph array, eg. a dot or a comma or an exclamation mark.. etc. In this case you don't need to perform any transformation on the character as it is not part of the alphabet.

Answer 2

The newAlph array already has the capitalized vowels.

The first if statement will add the letter from newAlph to the nstr variable if the current character is a letter.

Answer 3

As you already have answers to your question, here's a way to solve the problem with regular expressions and char codes:

function transform(str) {
  return str.replace(/[a-z]/gi, function(l) {
    var next = String.fromCharCode(l.charCodeAt(0) + 1);
    if (/[aeiou]/.test(next)) next = next.toUpperCase();
    return l == 'Z' ? 'A' : l == 'z' ? 'a' : next;
  });
}

transform('hello world'); //=> "Ifmmp xpsmE"

Answer 4

As @bagonyi stated, the else statement is for when a character in the string argument isn't found in the oldAlph array. In such a case this doesn't yield a negative number for n (as you speculated in your question), but rather yields undefined.

Furthermore, the code you posted doesn't solve for all possible strings, in particular those with capital letters already in them. So in Coderbyte when you run this code with the original test string "Argument goes here" in the Parameter Testing field, the output is "AshvnfOU hpft Ifsf"; notice that the "A" has not changed because it doesn't exist in oldAlph.