How to compare two string arrays, case insensitive and independent about ordering JavaScript, ES6

We need to write a function that compares two strings to check if they contain the same characters, ignoring case and order. This is useful for validating anagrams or checking string equivalence.

For example:

const first = 'Aavsg';
const second = 'VSAAg';
isEqual(first, second); // true

Using Array Sort Method

This method converts strings to arrays, sorts the characters, and compares the results. We extend the String prototype to add a sort method.

const first = 'Aavsg';
const second = 'VSAAg';

const stringSort = function(){
    return this.split("").sort().join("");
}

String.prototype.sort = stringSort;

const isEqual = (first, second) => first.toLowerCase().sort() === 
    second.toLowerCase().sort();

console.log(isEqual(first, second));

true

Using Character Frequency Map

This method counts character frequencies using a Map. Characters from the first string get +1, characters from the second string get -1. If all values sum to 0, the strings are equivalent.

const first = 'Aavsg';
const second = 'VSAAg';

const isEqual = (first, second) => {
    if(first.length !== second.length){
        return false;
    }
    
    first = first.toLowerCase();
    second = second.toLowerCase();
    const map = {};
    
    for(let ind = 0; ind  val === 0);
};

console.log(isEqual(first, second));

true

Using ES6 Map for Cleaner Code

Here's a more modern approach using ES6 Map and for...of loops:

const first = 'Aavsg';
const second = 'VSAAg';

const isEqual = (str1, str2) => {
    if (str1.length !== str2.length) return false;
    
    const charCount = new Map();
    
    // Count characters in first string
    for (const char of str1.toLowerCase()) {
        charCount.set(char, (charCount.get(char) || 0) + 1);
    }
    
    // Subtract characters from second string
    for (const char of str2.toLowerCase()) {
        if (!charCount.has(char)) return false;
        charCount.set(char, charCount.get(char) - 1);
    }
    
    // Check if all counts are zero
    return [...charCount.values()].every(count => count === 0);
};

console.log(isEqual(first, second));

true

Comparison of Methods

Conclusion

The Map-based approaches offer better time complexity O(n) compared to sorting O(n log n). Use the ES6 Map method for optimal performance and clean, readable code.