Solving "Missing letters" / freeCodeCamp Algorithm Challenges

Let's solve freeCodeCamp's intermediate algorithm scripting challenge, 'Missing letters'.

Starter Code

function fearNotLetter(str) {
  return str;
}

fearNotLetter("abce");

Instructions

Find the missing letter in the passed letter range and return it.

If all letters are present in the range, return undefined.

Test Cases

fearNotLetter("abce") should return "d".
fearNotLetter("abcdefghjklmno") should return "i".
fearNotLetter("stvwx") should return "u".
fearNotLetter("bcdf") should return "e".
fearNotLetter("abcdefghijklmnopqrstuvwxyz") should return undefined.

Our Approach

We have a short set of instructions for this one. After reading and looking at the test cases,

• We have one input, a string.
• We must return a string, compromised of a single letter, or undefined.
• We have to evaluate which letter is missing from the string, str. str is usually a string of continous (in alphabetical order) lowercased letters.

So, after looking at the test cases, the one which returns undefined is a string which contains every letter in the alphabet (in order). The instructions state, 'if all letters are present in the range', return undefined.

For this case, I thought we could handle that but doing a if statement to check if all the letters were contained. I just made a string of all letters, a-z.

if (str == 'abcdefghijklmnopqrstuvwxyz') return undefined;`

You could check the test cases with just this if statement and the undefined test case should pass. We still have a bit of work to do.

I am thinking we will have to make an array of the alphabet, a-z and also make an array from str and compare the two. Let's start off with that.

let alphabet = 'abcdefghijklmnopqrstuvwxyz'.split('');
console.log(alphabet)
// Array(26) [ "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", … ]

let str = "abcdefghjklmno";
let strArray = str.split('');
console.log(strArray);
// Array(14) [ "a", "b", "c", "d", "e", "f", "g", "h", "j", "k", … ]

So we have two arrays now. We can make a loop to check which one is missing but since strArray is just a certain amount of letters, we do not need to loop through the whole alphabet. I'll take the length of strArray to see how many times we have to execute the loop.

let strLen = strArray.length;

Since not all str (or strArray now) will begin with 'a', we have to check from where to start the comparison between the two arrays. The next variable I am going to create will show us where to begin. We are going to take the indexOf strArray[0] from alphabet. Have a look at the below example -

// Test Case: fearNotLetter("stvwx") should return "u".
str = "stvwx";
let strArray = str.split('');
let idx = alphabet.indexOf(strArray[0]);
let strLen = strArray.length;

// strArray = ['s', 't', 'v', 'w', 'x'];
// idx = 18; // 18 is the index of 's' in the alphabet
// strLen = 5;

So I think we have all the information we need to make the comparison. We have an alphabet array, a string array (to compare), the index to start the comparison at, and length to see how many times we need to run the loop to check.

To do this, I would like to do is run splice() on alphabet in order to get the exact part of the alphabet to compare.

Array.splice() (MDN)

let splicedAlphabet = alphabet.splice(idx, strLen);

So, we are exacting (from alphabet), the starting idx. For example

// Test Case: fearNotLetter("abce") should return "d".
alphabet = [ "a", "b", "c", "d", "e", "f", "g", "h", "i", "j", … ]
strArray = [ "a", "b", "c", "e" ]
let strLen = strArray.length; // 4
let idx = alphabet.indexOf(strArray[0]) // Checking where 'a' is, 0 index
let splicedAlphabet = alphabet.splice(idx, strLen);

// splicedAlphabet is taking (0, 4) // ['a', 'b', 'c', 'd']

So, with splicedAlphabet, we can check this against strArray to find the difference (or missing letter).

To execute the comparsion, I will use a for loop. Will run it on splicedAlphabet's length.

for (let i = 0; i < splicedAlphabet.length; i++) {
  if (!strArray.includes(splicedAlphabet[i])) return splicedAlphabet[i];
}

So, we are looking to see which letter is missing. We run an if statement within the for loop. If strArray doesn't include splicedAlphabet[i], that would be the missing letter, so we return it.

Array.includes() (MDN)

Our Solution

function fearNotLetter(str) {
  if (str == 'abcdefghijklmnopqrstuvwxyz') return undefined;

  let alphabet = 'abcdefghijklmnopqrstuvwxyz'.split('');
  let strArray = str.split('');
  let idx = alphabet.indexOf(strArray[0]);
  let strLen = strArray.length;
  let splicedAlphabet = alphabet.splice(idx, strLen);

  for (let i = 0; i < splicedAlphabet.length; i++) {
    if (!strArray.includes(splicedAlphabet[i])) 
      return splicedAlphabet[i]
  }
}

fearNotLetter("abce");

Links & Resources

'Missing letters' Challenge on fCC

freeCodeCamp

Donate to FCC!

Solution on my GitHub

Thank you for reading!

Comments

[Fred Kenneth Ekada Emojong]

`

`
function fearNotLetter(str) {
let alphabet='abcdefghijklmnopqrstuvwxyz';
let index= alphabet.indexOf(str[0]);
let realOrder=alphabet.slice(index,index+str.length);
let missing="";

for(let letters of realOrder){
// !str.includes(letters)?missing+=letters: null;
!str.includes(letters)?missing+=letters: null;
}

return missing || undefined;
}

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

[Abhishek Prasad]

function fearNotLetter(str) {
let dummy = 'abcdefghijklmnopqrstuvwxyz'
let x = dummy.slice(dummy.indexOf(str[0]))
let s='';

for(let j=0;j<x.length;j++){
if(str.slice(-1) == x.charAt(j)){
s = s.concat(x.charAt(j))
break;
}
else {
s = s.concat(x.charAt(j))
}
}

return s[[...str].findIndex((el, index) => el !== s[index])]
}

fearNotLetter("abce");

[Abhishek Prasad]

i have done it like this do you have any suggestions ....