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Anagrams Checker - Three JavaScript Solutions

sarah_chima profile image Sarah Chima ・4 min read

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. Earlier this week, I wrote an article on the Big O notation. If you are not familiar with it, you might want to read it as some concepts are used in this article. Let's go straight to the problem statement.

Finding Anagrams - the problem

Anagrams are words that have the same characters in the same quantity. This means that two strings are anagrams if we can rearrange one to get the other.

Here are some examples of words that are anagrams.

  1. “listen” and “silent”
  2. “rail safety” and “fairy tales”
  3. “dormitory” and “dirty room”
  4. “the eyes” and “they see”

To solve this problem, we will assume the following:

  1. That we ignore extra characters like “!”, “@”, etc. and whitespaces.
  2. We only want to work with lowercase characters.

Let us look at some solutions to this problem. Then we'll compare each of them based on their time complexity.

Solution 1 - Create a character map of both strings and compare maps

A character map in this context is a map or object that contains each unique character in the string. It stores the character as a key and the number of times it occurs in that string as the value.

    function anagrams(stringA, stringB) {
        /*First, we remove any non-alphabet character using regex and convert
        convert the strings to lowercase. */
        stringA = stringA.replace(/[^\w]/g, "").toLowerCase()
        stringB = stringB.replace(/[^\w]/g, "").toLowerCase()

        //Get the character map of both strings
        const charMapA = getCharMap(stringA)
        const charMapB = getCharMap(stringB)

        /* Next, we loop through each character in the charMapA, 
        and check if it exists in charMapB and has the same value as
        in charMapA. If it does not, return false */
        for (let char in charMapA) {
            if (charMapA[char] !== charMapB[char]) {
                return false
            }
        }

        return true
    }

    function getCharMap(string) {
        // We define an empty object that will hold the key - value pairs.
        let charMap = {}

        /*We loop through each character in the string. if the character 
        already exists in the map, increase the value, otherwise add it 
        to the map with a value of 1 */
        for (let char of string) {
            charMap[char] = charMap[char] + 1 || 1
        }
        return charMap
    }

The runtime complexity of a for loop is linear i.e O(n). In this case, there are 3 consecutive forloops which are not nested. Ignoring constants and other factors, the time complexity is approximately linear i.e. O(n).

2. Sort Strings and check if they are the same

This is a shorter and neater way of checking if two strings are anagrams.
In this case, we convert the string to an array, use the Array.sort()
method to sort it and convert it back to a string. Then we compare both strings and check if they are the same.

    function anagrams(stringA, stringB) {
        /*First, we remove any non-alphabet character using regex and convert       
        convert the strings to lowercase. */
        stringA = stringA.replace(/[^\w]/g, '').toLowerCase()
        stringB = stringB.replace(/[^\w]/g, '').toLowerCase()

        return sortString(stringA) === sortString(stringB)
    }

    /*This function sorts the strings*/ 
    function sortString(string) {
        return string.split('').sort().join('');
    }

Array.sort uses merge sort so its time complexity is O(nlogn).

3. Using Array.splice()

This is yet another solution. In this case, we convert string B to an array, loop through each character in string A and check if it exists in an array of string B, arrB. If it exists, we use the Splice method to remove it from the array. We do this so that characters that occur more than once in arrB are not checked twice.

    function anagrams(stringA, stringB) {
        /*First, we remove any non-alphabet character using regex and convert       
        convert the strings to lowercase. */
        stringA = stringA.replace(/[^\w]/g, '').toLowerCase()
        stringB = stringB.replace(/[^\w]/g, '').toLowerCase()

        /*Next, we check if the lengths of the strings are equal. 
        If they are anagrams, they will have the same length. */
        if (stringA.length !== stringB.length) {
            return false
        }

        let arrB = stringB.split("")

        for (let char of stringA ){ 
            if (!arrB.includes(char)) {
                return false
                break;
            } else {
                arrB.splice(arrB.indexOf(char), 1)
            }
        }

        return true

    }

So let's consider the time complexity of this solution. In this case, there are three loops that run. The for loop, the includes loop and the splice loop. Since the splice loop and the includes are not nested, the time complexity tends to O(n^2 ).

Conclusion

We have seen the solutions and their approximate time complexities. Comparing their time complexities, the first solution seems to have better performance. It has an approximate time complexity of O(n). The second solution, however, is more concise. So you can choose any solution depending on what is more important to you.

Got any question or addition? Please leave a comment.

Thank you for reading.

Discussion (31)

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codemouse92 profile image
Jason C. McDonald • 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.

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aadibajpai profile image
Aadi Bajpai

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

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codemouse92 profile image
Jason C. McDonald • 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.

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codemouse92 profile image
Jason C. McDonald • 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.

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aadibajpai profile image
Aadi Bajpai

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').

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codemouse92 profile image
Jason C. McDonald

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

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evanoman profile image
Evan Oman

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.

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aadibajpai profile image
Aadi Bajpai

Not very efficient or fast, however.

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evanoman profile image
Evan Oman

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

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aadibajpai profile image
Aadi Bajpai

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

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codemouse92 profile image
Jason C. McDonald • 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.

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calebpitan profile image
Caleb Adepitan • 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"))

Copyright (c) 2019 Caleb Pitan
😇😁😁

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lexlohr profile image
Alex Lohr

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

anagrams('abc', 'aad') // => true
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calebpitan profile image
Caleb Adepitan

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.

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joeberetta profile image
Joe Beretta

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

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calebpitan profile image
Caleb Adepitan

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

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joeberetta profile image
Joe Beretta

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

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moopet profile image
Ben Sinclair

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

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calebpitan profile image
Caleb Adepitan • 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.

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moopet profile image
Ben Sinclair

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.

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calebpitan profile image
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sem8 profile image
Sem Limi • 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;
  }
Enter fullscreen mode Exit fullscreen mode

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

console.log(anagrams("anagram", "nagarams")); // false
Enter fullscreen mode Exit fullscreen mode
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lexlohr profile image
Alex Lohr

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);
}
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pavelloz profile image
Paweł Kowalski

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 :)

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_bigblind profile image
Frederik 👨‍💻➡️🌐 Creemers

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
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dksami profile image
Raja Osama • Edited

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

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joeattardi profile image
Joe Attardi

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

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lauragift21 profile image
Gift Egwuenu

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

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sarah_chima profile image
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shivraj97 profile image
Shivraj97

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

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alexpanchuk profile image
Alex Panchuk

For example?