Simplest code to check if a string contains all the alphabet using JavaScript

Question

This is the simplest solution I found. It returns true or false for the given string if it contains all letters of the alphabet in it or not.

Here is the code I found:

new Set("A quick brown fox jumps over the lazy dog"
  .toLowerCase()
  .replace(/[^a-z]/gi, "")
  .split("")
).size === 26

Any other simpler form of checking to see if a string contains all of the letters in the alphabet would be helpful.

Thanks!

Answer 1

You don't need to split

As it would turn out, you don't need to run String#split before passing your string to new Set. The constructor for Set, when passed a string, will, essentially, split it into single characters for you before creating the set.

Example:

new Set('A quick brown fox jumps over the lazy dog'
  .toLowerCase()
  .replace(/[^a-z]/g, '')
).size === 26

This works just as well because something like new Set('test') turns into

Set(3) {"t", "e", "s"}

On a side note, you can see that I've removed the i flag from the regular expression as pointed out by one of the other answers as it is unnecessary due to the .toLowerCase()

Answer 2

You can avoid the regex and also return early from the function once you have all the letters with something like this. It creates a set of all the letters and removes them as you find them. Once the set is empty you can return. If the loop finishes, you didn't remove everything. This only requires space for the alphabet set and since set operations are constant time, this is O(n) in the worst case.

function allLetters(str){
    let alpha = new Set("abcdefghijklmnopqrstuvwxyz")
    for (let c of str.toLowerCase()){
        alpha.delete(c)
        if (alpha.size == 0) return true
    }
    return false
}

let text = "hello my name if gunther"
let text2 = "The quick brown fox jumps over the lazy dog"

console.log(allLetters(text))
console.log(allLetters(text2))

Answer 3

This is the simplest code I found, It returns true or false for the given string mentioning the string contains all the alphabet in it or not.

new Set("<your_string>".toLowerCase().replace(/[^a-z]/g, "") ).size === 26

Example:

new Set("A quick brown fox jumps over the lazy dog".toLowerCase().replace(/[^a-z]/g, "") ).size === 26

Any other simplest form of code can be helpful. Please share it.

Thanks!

Answer 4

I believe this is the "simplest" w.r.t. computational complexity, requiring O(1) space (to store the character frequency table, assuming a fixed upper-bound possible input alphabet) and O(n) time as it iterates over the input string only once (plus a constant-time for the final check over the alphabet string).

var inputString = "Jaded zombies acted quaintly but kept driving their oxen forward";

var charCounts = {};
for( var i = 0; i < inputString.length; i++ ) {
    var c = inputString.at( i ).toLower();
    if( charCounts[c] ) charCounts[c]++;
    else                charCounts[c] = 1;
}

var alphabet = "abcdefghijklmnopqrstuvwyz";
for( var a = 0; a < alphabet.length; a++ ) {
    if( !charCounts[ alphabet.at(a) ] ) {
        console.log( "char %s does not appear in input string.", alphabet.at(a) );
    }
}

Answer 5

As I look at it again, I can provide one tiny improvement to your code:

new Set("<your_string>".toLowerCase().replace(/[^a-z]/g, "").split("")).size === 26 .

^{Remove the 'i' flag on the regex because it's lowercased.}

Answer 6

Here is a different way to due it using String.fromCharCode() and every()

const allLetters = (str) => Array.from({length:26}).map((x,i) => String.fromCharCode(i+97)).every(a => str.toLowerCase().includes(a));
console.log(allLetters("abcdefghijklmnopqrstuvwxyz"));

Or you can hardcode all the alphabets.

const allLetters = (str) => [..."abcdefghijklmnopqrstuvwxyz"].every(x => str.toLowerCase().includes(x));

console.log(allLetters('abcdefghijklmnopqrstuvwxyz'))
console.log(allLetters('abcdefghijklmnopqyz'))

Answer 7

function isPangram(sentence){
   let lowerCased = sentence.toLowerCase();
   for(let char of 'abcdefghijklmnopqrstuvwxyz'){
      if(!lowerCased.includes(char)){
         return false
       }
    }
  return true
}

Here is another way using a for...of loop.