Intermediate Algorithm Scripting: Missing letters - Help Needed with Converting Letters to Char Code and Back

[DragonOsman]

I tried to do this by looping through a range of letters converted to their ASCII codes, with the range specified by the starting and ending letters of the given string.

But the way I wrote the function gives “undefined”. When I console.log str.codePointAt(endLetter)), I get “97” which is the same as the one for startLetter and I don’t get why. I tried it charCodeAt as well and had the same result.

Would someone please tell me what I’m doing wrong here?

My code so far

function fearNotLetter(str) {
const startLetter = str[0];
const endLetter = str[str.length - 1];
let missingLetter;

for (let i = str.codePointAt(startLetter); i < str.codePointAt(endLetter); ++i) {
  if (!str.includes(String.fromCharCode(i))) {
    missingLetter = String.fromCharCode(i);
  }
}

return missingLetter;
}

console.log(fearNotLetter("abce"));

My browser information:

User Agent is: Mozilla/5.0 (Windows NT 10.0; Win64; x64) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/84.0.4147.89 Safari/537.36 Edg/84.0.522.40.

Challenge: Missing letters

Link to the challenge:

freeCodeCamp.org

[snigo]

@DragonOsman

function fearNotLetter(str) {
  const startLetter = str[0]; // You don't need letter, but a code point of the letter str.codePointAt(0);

  const endLetter = str[str.length - 1]; // Unnecessary line

  let missingLetter; // Unnecessary line

  // startLetter would already be a code point
  // iterating until startLetter + str.length would have same effect
  // ++i will not have any effect here, if you wanted to start from the second letter, then it should have been done in i initialization part
  for (let i = startLetter + 1;  i < startLetter + str.length; i++) {
    if (!str.includes(String.fromCharCode(i))) {
      missingLetter = String.fromCharCode(i); // Congrats, you have found it. No need to iterate further and you can just return from here
    }
  }

  return missingLetter;
}

Hope this would help

[snigo]

Digging a bit further and becoming more picky, your code doesn’t have to have quadratic complexity, meaning you don’t really need to search string for character inside for loop. Think if you can do it without str.includes

[ccarri11]

DragonOsman right on!!

Just another approach to solve if you get stuck:

Try converting the values first into numbers and storing those into an array, and just use your counter i to start from 0 to length-1 as you traverse each letter.

Example: “abcd” =

for(let i = 0; i < str.length;i++){
// charCodeAt()
}

[number values 1, number values 2]…

Then convert those numbers from the array to strings and concat each letter with empty str, returning the result.

[DragonOsman]

So you mean startLetter should be the code for the letter to start with? And how will I do it without str.includes?

[DragonOsman]

Also, why does the loop have to go from startLetter + 1 to startLetter + str.length? I need to understand that. Thanks.

[snigo]

Never in your code you use actual first letter, but the code point of the first letter. Why would you first extract first letter and then extract code of that letter. One unnecessary action.

It’s fairly easy to drop .includes as you sequentially iterating until the end of the string.

You don’t need to check the first letter, right? Why not staring with the second right away.