why doesn't this code work for anagram checking?

Question

I'm trying make an anagram check in javascript. For simplicity, assume that the function below only takes lowercase strings without any spacing/numbers/symbols. Why doesn't the code below work?

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

score 5 · Answer 1 · answered Jan 10 '16 at 02:30

The == does not work for Array, as Array is an Object. The == operator checks if the Objects are the SAME:

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

Thus, the simplest was to check this is to convert them back into String and use ==, since == does work with String:

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

score 2 · Answer 2 · edited May 23 '17 at 11:44

2

You can't compare array elements directly like that in Javascript.

Here are multiple implementations that will do that: How to check if two arrays are equal with JavaScript?.

In your case, you could simply use a .join() to compare two strings instead:

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");
    }
}

edited May 23 '17 at 11:44

Community

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answered Jan 10 '16 at 02:29

Mr_Pouet

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thanks. did not realize that arrays can't be compared. – d1du Jan 10 '16 at 02:33

score 0 · Answer 3 · answered Jan 10 '16 at 03:47

I'm trying make an anagram check in javascript.

Well this is not exactly an Answer to this question but a different Approach to this Problem; and in my first tests, way faster (33-50%).
Mr_Pouet and Andrew Templeton already gave you an answer to this exact question.

function anagram(a, b, caseSensitive){
    if(a.length !== b.length) return false;
    var c = caseSensitive? a: a.toLowerCase(), 
        d = caseSensitive? b: b.toLowerCase(),
        map = new Array(128),
        cc;

    for(var i = a.length; i--; ){
        cc = c.charCodeAt(i);
        map[cc] = (map[cc] || 0) + 1;

        cc = d.charCodeAt(i);
        map[cc] = (map[cc] || 0) - 1;
    }

    var i = map.length;
    if(i < 256){
        //faster for ANSI-like text
        while(i--) 
            if(i in map && map[i]) return false;
    }else{
        //faster for a huge sparse map, 
        //like UTF8 multibytes (Asian characters for example)
        for(var k in map)
            if(map[k]) return false;
    }
    return true;
}

why doesn't this code work for anagram checking?

3 Answers3