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Anagrams Checker - Three JavaScript Solutions

Sarah Chima on August 09, 2019

I started a series on JavaScript solutions to common algorithms and javascript problems. In case you missed the first one, here's a link to it. Ear...

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Jason C. McDonald • Aug 9 '19 • Edited

It's Python (so, off-topic), but I couldn't help myself:

def anagram_checker(word1, word2):
    word1 = [c for c in word1.lower() if c.isalpha()]
    word2 = [c for c in word2.lower() if c.isalpha()]
    return (len(word1) == len(word2) and set(word1) == set(word2))

O(n), if you're wondering.

Aadi Bajpai • Aug 11 '19

Flawed, would fail for strings like ('aabb', 'aaab')

Jason C. McDonald • Aug 11 '19 • Edited

Good catch. I'd overlooked that edge case. Let me ponder....

AHA! The most obvious solution is...

def anagram_checker(word1, word2):
    word1 = sorted([c for c in word1.lower() if c.isalpha()])
    word2 = sorted([c for c in word2.lower() if c.isalpha()])
    return word1 == word2

sorted() employs Timsort, which has a worst case of O(n log n). Everything else is still only O(n).

Naturally, I'm sure there's a more efficient solution if I chew on this longer, but that'd be the one I'd probably use in production; reasonable performance, readable, maintainable, etc.

Jason C. McDonald • Aug 11 '19 • Edited

EVEN BETTER! I just thought of a completely O(n) solution that relies on the commutative property of addition, and the fact that any letter would have a unique numeric equivalent (retrieved via ord()).

def anagram_checker(word1, word2):
    val1 = sum([ord(c) for c in word1.lower() if c.isalpha()])
    val2 = sum([ord(c) for c in word2.lower() if c.isalpha()])
    return val1 == val2

Ahhh, I love a good hack.

Aadi Bajpai • Aug 11 '19

Nope, this is flawed too. There are multiple ways to express a sum. 3+2=5 and 4+1=5 as well.

So ord('a') + ord('z') = ord('b') + ord('y')

Would fail for strings such as ('az', 'by').

Jason C. McDonald • Aug 11 '19

Hmm, hadn't considered that angle....

Evan Oman • Aug 12 '19

Prime factorizations are unique so if you map each letter to a prime and then take the product of those primes, you'd have a unique representation of the characters in the word. See this classic tweet for a better explanation.

Aadi Bajpai • Aug 13 '19

Not very efficient or fast, however.

Evan Oman • Aug 13 '19

I mean its time complexity is O(n)-ish but yeah the large number multiplication would get you for longer words.

Aadi Bajpai • Aug 13 '19

26th prime is 101 so I imagine zzzzzzzzzzzzz et all would take more time

Jason C. McDonald • Oct 2 '19 • Edited

It's not efficient, but I wonder if a Cantor pairing of the ord() values would work. It would certainly be optimal if all the strings were two letters. The trouble is, those numbers get very big, very fast.

Caleb Adepitan • Aug 9 '19 • Edited

What do you think?

function anagrams(a, b) {
  let i = 0
  let sumA = 0
  let sumB = 0
  a = a.replace(/[\W]/g, "").toLowerCase()
  b = b.replace(/[\W]/g, "").toLowerCase()
  if (a.length !== b.length) return false
  !(function sumCharCode() {
    if (i < a.length) {
      sumA += a.charCodeAt(i)
      sumB += b.charCodeAt(i)
      i++
      sumCharCode()
    }
  })()
  return sumA === sumB;
}

console.log(anagrams("listen", "silent"))

Ben Sinclair • Aug 10 '19

I think that function expressions using ! are there just to make the code less readable.

Caleb Adepitan • Aug 10 '19 • Edited

My bad! I won't lie to you I really don't know the use of the "!", but I think it's to indicate an iife inside the function, hence aid readability. I put it because in my early years I saw a lot of these around, and I thought having it around or not affects nothing. Didn't know it affects readability, although I don't understand how. I wouldn't mind if you told me.

Ben Sinclair • Aug 10 '19

I prefer explicit to implicit (comes from enjoying Python too much...) and !(function(){...}(); is just... well, if you haven't seen it before you have no idea what the gosh darnit is going on.

Caleb Adepitan • Aug 10 '19

digitalfortress.tech/js/exclamatio...

Alex Lohr • Aug 12 '19

I initially thought about that approach, but discarded it for an obvious reason:

anagrams('abc', 'aad') // => true

Caleb Adepitan • Aug 12 '19

Smart 😃.
It's just too obvious. Didn't take time to analyze it. It struck my mind as I saw this post and I put it down.

Joe Beretta • Aug 9 '19

I thought about using \W instead of ^w as you wrote)

Caleb Adepitan • Aug 9 '19

There's really no difference, it's just left to choice.

Joe Beretta • Aug 9 '19

Know it. But I think this way is more elegant and perfomance

Sem Limi • Jul 15 '20 • Edited

Solution 1 might not be completely correct if you don't check that the length of s and t are the same otherwise return false like so:

 if (s.length !== t.length) {
    return false;
  }

Otherwise this test will return true even though it should be false

console.log(anagrams("anagram", "nagarams")); // false

Frederik 👨‍💻➡️🌐 Creemers • Oct 2 '19

Here's an O(∞) solution inspired by bogosort. If it halts, a and b are anagrams.

def anagrams(a, b):
    while True:
        b = str(random.shuffle(b))
        if a == b:
            return True

Alex Lohr • Aug 12 '19

My first solution (ES2019) looks like this:

const isAnagram = (word1, word2) => {
    const charsInWord = {};
    const maxLength = Math.max(word1.length, word2.length);
    for (let index = 0; index < maxLength; index++) {
        const char1 = word1.charCodeAt(index);
        const char2 = word2.charCodeAt(index);
        if (char1 >= 97 && char1 <= 122) {
            charsInWord[char1] = (charsInWord[char1] || 0) + 1;
        }
        if (char2 >= 97 && char2 <= 122) {
            charsInWord[char2] = (charsInWord[char2] || 0) - 1;
        }
    }
    return (Object.values(charsInWord) || { a: 0 }).every((charCount) => charCount === 0);
}

Paweł Kowalski • Aug 10 '19

My intuition, before i saw any of the solutions, went into the 2nd one because it feels like it would be the easiest to understand and the fastest. I wonder if thats the case ;)
Thank you for the brain teaser, good beginning of the day :)

Developeranees • May 5 '22

Adams Stark • Mar 14 '21

In order to remove spaces, you are calling string.replace once for every occurrence of " " (space) in each string. An easier way to remove whitespace in a string is to use a regular expression object with a global modifier, which will replace all matching characters in the string. Then, you can get rid of your while-loops, and your code becomes slightly easier to read. I hope it will help you to complete your Anagram checker project .

  ## function findAnagram (firstWord, secondWord) {
// "/ /g" is a regular expression object that finds all spaces in a string.
secondWord = secondWord.replace(/ /g, "");
firstWord = firstWord.replace(/ /g, "");
...

 ###

You can also use the /\s/g regular expression object to replace all whitespace including tabs, newlines, etc.

Raja Osama • Jan 20 '21 • Edited

tried all the above and the below solutions but the least execution time was taken by this one.

Gift Egwuenu • Aug 9 '19

Great article Sarah! Mind sharing any resources for leveling up on Data Structures and Algorithms. Thanks

Sarah Chima • Aug 15 '19

I'm taking a course by Stephen Grider on Udemy. Another recommended course is by Colt Steele - JavaScript Algorithms and Data Structures Masterclass.

Joe Attardi • Aug 9 '19

This is a favorite interview question of mine. The character map solution is what most people seem to do - myself included!

Shivraj97 • Jul 21 '20

Solution 2 is not correct. What if I pass a word with extra letters to string 2???

Alex Panchuk • Oct 24 '20

For example?