What is the reason behind the failure of this code to check for anagrams?

Question

What is the reason behind the failure of this code to check for anagrams?

I am attempting to create a JavaScript function that checks for anagrams. To keep things simple, let's assume this function only works with lowercase strings that do not contain spaces, numbers, or symbols. Why isn't the following code functioning properly?

var checkAnagram = function(word1, word2) {
    var arrayWord1 = word1.split('').sort();
    var arrayWord2 = word2.split('').sort();
    if (arrayWord1 == arrayWord2) {
        console.log("These words are anagrams");
    }
    else {
        console.log("These words are not anagrams");
    }
}

javascript arrays sorting split anagram

Answer 1

Answer №1

When comparing Array, the == operator does not work because an Array is considered an Object. The == operator checks if the two objects are the same:

var foo = {};
var bar = {};
console.log(foo == bar); // false
var foo2 = {};
var bar2 = foo2;
console.log(foo2 == bar2); // true

To compare arrays, a simple solution is to convert them into strings and then use the == operator since it works with String:

var anagram = function(string1, string2) {
    var string1array = string1.split('').sort();
    var string2array = string2.split('').sort();
    // Use .join('') on both to compare.
    if (string1array.join('') == string2array.join('')) {
        console.log("they're anagrams");
    }
    else {
        console.log("they are not anagrams");
    }
}

Answer 2

When comparing Array, the == operator does not work because an Array is considered an Object. The == operator checks if the two objects are the same:

var foo = {};
var bar = {};
console.log(foo == bar); // false
var foo2 = {};
var bar2 = foo2;
console.log(foo2 == bar2); // true

To compare arrays, a simple solution is to convert them into strings and then use the == operator since it works with String:

var anagram = function(string1, string2) {
    var string1array = string1.split('').sort();
    var string2array = string2.split('').sort();
    // Use .join('') on both to compare.
    if (string1array.join('') == string2array.join('')) {
        console.log("they're anagrams");
    }
    else {
        console.log("they are not anagrams");
    }
}

Answer 3

Answer №2

In the realm of Javascript, direct comparison of array elements is not possible.

If you are seeking ways to compare two arrays in JavaScript, here's a resource that provides multiple implementations: Ways to Compare Arrays in JavaScript.

For your current scenario, a simple solution would be to utilize the .join() method to compare two strings as shown below:

var anagram = function(string1, string2) {
    var string1_sorted = string1.split('').sort().join('');
    var string2_sorted = string2.split('').sort().join('');
    if (string1_sorted == string2_sorted) {
        console.log("They're anagrams");
    } else {
        console.log("They are not anagrams");
    }
}

Answer 4

In the realm of Javascript, direct comparison of array elements is not possible.

If you are seeking ways to compare two arrays in JavaScript, here's a resource that provides multiple implementations: Ways to Compare Arrays in JavaScript.

For your current scenario, a simple solution would be to utilize the .join() method to compare two strings as shown below:

var anagram = function(string1, string2) {
    var string1_sorted = string1.split('').sort().join('');
    var string2_sorted = string2.split('').sort().join('');
    if (string1_sorted == string2_sorted) {
        console.log("They're anagrams");
    } else {
        console.log("They are not anagrams");
    }
}

Answer 5

Answer №3

In my JavaScript coding journey, I have embarked on creating an efficient anagram check.

Although not a direct response to the initial query, here is an alternate method for tackling this issue; one that has proven to be notably quicker in my preliminary evaluations (by 33-50%).
Both Mr_Pouet and Andrew Templeton have already provided solutions for the original question.

function checkAnagram(wordA, wordB, isSensitive){
    if(wordA.length !== wordB.length) return false;
    var a = isSensitive? wordA: wordA.toLowerCase(), 
        b = isSensitive? wordB: wordB.toLowerCase(),
        dictionary = new Array(128),
        charCode;

    for(var j = wordA.length; j--; ){
        charCode = a.charCodeAt(j);
        dictionary[charCode] = (dictionary[charCode] || 0) + 1;

        charCode = b.charCodeAt(j);
        dictionary[charCode] = (dictionary[charCode] || 0) - 1;
    }

    var index = dictionary.length;
    if(index < 256){
        //optimized for ASCII characters
        while(index--) 
            if(index in dictionary && dictionary[index]) return false;
    }else{
        //suited for extensive sparse mappings, such as UTF8 multibytes
        for(var key in dictionary)
            if(dictionary[key]) return false;
    }
    return true;
}

Answer 6

In my JavaScript coding journey, I have embarked on creating an efficient anagram check.

Although not a direct response to the initial query, here is an alternate method for tackling this issue; one that has proven to be notably quicker in my preliminary evaluations (by 33-50%).
Both Mr_Pouet and Andrew Templeton have already provided solutions for the original question.

function checkAnagram(wordA, wordB, isSensitive){
    if(wordA.length !== wordB.length) return false;
    var a = isSensitive? wordA: wordA.toLowerCase(), 
        b = isSensitive? wordB: wordB.toLowerCase(),
        dictionary = new Array(128),
        charCode;

    for(var j = wordA.length; j--; ){
        charCode = a.charCodeAt(j);
        dictionary[charCode] = (dictionary[charCode] || 0) + 1;

        charCode = b.charCodeAt(j);
        dictionary[charCode] = (dictionary[charCode] || 0) - 1;
    }

    var index = dictionary.length;
    if(index < 256){
        //optimized for ASCII characters
        while(index--) 
            if(index in dictionary && dictionary[index]) return false;
    }else{
        //suited for extensive sparse mappings, such as UTF8 multibytes
        for(var key in dictionary)
            if(dictionary[key]) return false;
    }
    return true;
}

What is the reason behind the failure of this code to check for anagrams?

Answer №1

Answer №2

Answer №3

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