Find missing letter in list of alphabets

Question

I am trying to solve the following issue:

Find the missing letter in the passed letter range and return it. If all letters are present in the range, return undefine

Answer 1

Note that you have a typo in alphabet: There are two "e"s.

You could split the string into an array, then use the some method to short-circuit the loop when you don't find a match:

function fearNotLetter(str) {
  var alphabet = 'abcdefghijklmnopqrstuvwxyz',
      missing,
      i= 0;
  
  str.split('').some(function(l1) {
    var l2= alphabet.substr(i++, 1);
    if(l1 !== l2) {
      if(i===1) missing= undefined;
      else      missing= l2;
      return true;
    }
  });
  
  return missing;
}

console.log(fearNotLetter('abce'));             //d
console.log(fearNotLetter('bcd'));              //undefined
console.log(fearNotLetter('abcdefghjklmno'));   //i
console.log(fearNotLetter('yz'));               //undefined

Answer 2

  function fearNotLetter(str) {
  //transform the string to an array of characters    
  str = str.split('');

  //generate an array of letters from a to z
  var lettersArr = genCharArray('a', 'z');
  //transform the array of letters to string
  lettersArr = lettersArr.join('');
  //substr the a to z string starting from the first letter of the provided 
  string
  lettersArr = lettersArr.substr(lettersArr.indexOf(str[0]), str.length);
  //transform it again to an array of letters
  lettersArr = lettersArr.split('');
  //compare the provided str to the array of letters
  for(var i=0; i<lettersArr.length;i++){
    if(str[i] !== lettersArr[i])
      return lettersArr[i];
  }
  return undefined;
}

function genCharArray(charA, charZ) {
    var a = [], i = charA.charCodeAt(0), j = charZ.charCodeAt(0);
    for (; i <= j; ++i) {
        a.push(String.fromCharCode(i));
    }
    return a;
}

fearNotLetter("bcd");

Answer 3

This will do what you're looking for:

Hit run and check your console

function missingLetter (str) {
    var alphabet = ("abcdefghijklmnopqrstuvwxyz");
    var first = alphabet.indexOf(str[0]);
    var strIndex = 0;
    var missing;

    for (var i = first ; i < str.length ; i++) {
        if (str[strIndex] === alphabet[i]) {
            strIndex++;
        } else {
            missing = alphabet[i];
        }
    }
    return missing;
}

console.log(missingLetter("abce"));
console.log(missingLetter("bcd"));
console.log(missingLetter("abcdefghjklmno"));
console.log(missingLetter("yz"));

Answer 4

Try this:

function fearNotLetter(str) {
    var alp = ('abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ').split(''), i;
    for (i =  alp.indexOf(str.charAt(0)); i < str.length; i++) {
        if (str.split('').indexOf(alp[i]) === -1) {
            return alp[i];
        }
    }
    return undefined;
}

fearNotLetter('bcd');

Answer 5

Here's my solution, which i find to be quite easy to interpret:

function fearNotLetter(str) {
  var missingLetter;
  for (var i = 0; i < str.length; i++) {
    if (str.charCodeAt(i) - str.charCodeAt(i-1) > 1) {
      missingLetter = String.fromCharCode(str.charCodeAt(i) - 1);
    }
  }
  return missingLetter;
}

Answer 6

I just did this challenge. I am a beginner to Javascript so this was my approach, very simple, sometimes you don't have to use the methods they provide but they also help. I hope you can understand it.

function fearNotLetter(str) {

var alphabet= "abcdefghijlmnopqrstuvwxyz";
var piece =alphabet.slice(0, str.length+1);




for(var i=0; i < piece.length; i++ ){
if(str.charCodeAt(0) != 97){

return undefined;
}

else if(str.indexOf(piece[i])===-1){
  return piece[i];
}
}// for loop

}

fearNotLetter("abce");// It will return d
fearNotLetter("xy");//It will return undefined
fearNotLetter("bce");//It will return undefined