Help needed on Missing Letters

JavaScript
#1

I’m solving Intermediate Algorithm scripting problems and I have to find the missing letter passed in the letter range and return them. If all letters are present, undefined must be returned.

 function fearNotLetter(str) {
  var alph = "abcdefghijklmnopqrstuvwxyz";
  var newstr;
  for(var i = 0; i < alph.length; i++){
    if(str[i] !== alph[i])
    {
newstr = alph[i];
break;
    }
  }
  return newstr;
}

fearNotLetter("abce");

So far my code is working fine for the first two and last conditions. But it’s returning false/wrong for the middle two conditions. Kindly please help and point where I’m doing wrong.

#2

what should you do if str[0] is not an "a"?

your function is returning "a" if that is the case

#3

But I didn’t specifically define str[0] to be “a”. How does this work?

#4

but alph[0] is always “a”, so when the check str[0] !== arr[0] is made and str starts somewhere else in the alphabet, this happen

look at the failing tests
fearNotLetter("stvwx")

in this case you have str[0] !== alph[0] being true because str[0] is “s”, and the function returns “a”

#5

Can you tell me how to rectify this problem? Should I split “alph”?

#6

consider this, if the string doesn’t always start from a, you can’t always check the alphabet from the a - there are a few ways to make this work, try thinking of a way

#7

Can I use “Includes” method?

#8

you could, but I honestly can’t think of a way how. if you have an idea, try!

#9

I solved it this way.

 function fearNotLetter(str) {
  var alph = "abcdefghijklmnopqrstuvwxyz";
  var alph = alph.substr(alph.indexOf(str[0]), str.length);
  var newstr;
 for(var i = 0; i < alph.length; i++){
    if(str[i] !== alph[i]){
      newstr = alph[i];
      break;
      }
    }  
return newstr;
}

fearNotLetter("abce");
#10