Intermediate Algorithm Scripting - Missing letters

[benomarlaid7]

Tell us what’s happening: i didnt understand whats means { return String.fromCharCode(charCode - 1); } *
Describe your issue in detail here.
thanks
**
function fearNotLetter(str) {
for (let i = 0; i < str.length; i++) {
/ code of current character */
const charCode = str.charCodeAt(i);

/* if code of current character is not equal to first character + no of iteration
    then a letter was skipped */
if (charCode !== str.charCodeAt(0) + i) {
  /* if current character skipped past a character find previous character and return */
  return String.fromCharCode(charCode - 1);
}

}
return undefined;
}

// test here
fearNotLetter(“abce”);**

function fearNotLetter(str) {


 for (let i=0; i<str.length; i++) {
   const charCode = str.charCodeAt(i);
    
   if(charCode !== str.charCodeAt(0)+ i) {
     
     return String.fromCharCode(charCode - 1);
     
   })
    }
      return undefined;
}


fearNotLetter("abce");
//d
//abce

//0<4 97 = 97    97 !== 101   97+3 =100  97+4 101

//1     98 
//2       99
//3     101

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Challenge: Intermediate Algorithm Scripting - Missing letters

Link to the challenge:

freeCodeCamp.org

[bvanb]

I assume that you are looking at an example answer. This answer relies on knowledge of character codes. The lowercase letters abc...z are represented by the numbers 97 to 122. Therefore, a set of consecutive letters will correspond to a set of consecutive character code numbers.

If you have the string "abce", the function will find the character code of the first letter (97), then check if every consecutive letter has a character code one higher than the last. It will find the following:

a | 97
b | 98
c | 99
e | 101

When it gets to 101, it sees that a letter has been skipped. Now, it has to return the missing letter, which will have a character code one less than the letter that was found. Therefore, it subtracts 1 from 101 and converts the character code into the letter d by calling String.fromCharCode(charCode - 1).