Finding and returning uncommon characters between two strings in JavaScript

Problem

We are required to write a JavaScript function that takes in two strings. Our function should return a new string of characters which is not common to both the strings.

Example

Following is the code −

const str1 = "xyab";
const str2 = "xzca";
const findUncommon = (str1 = '', str2 = '') => {
    const res = [];
    for (let i = 0; i < str1.length; i++){
        if (!(str2.includes(str1[i]))){
            res.push(str1[i])
        }
    }
    for (let i = 0; i < str2.length; i++){
        if (!(str1.includes(str2[i]))){
            res.push(str2[i])
        }
    }
    return res.join("");
};
console.log(findUncommon(str1, str2));

Output

ybzc

How It Works

The function compares each character from both strings:

• First loop: Finds characters in str1 that don't exist in str2
• Second loop: Finds characters in str2 that don't exist in str1
• Returns concatenated uncommon characters

Alternative Approach Using Set

A more efficient approach using Set for faster lookups:

const findUncommonWithSet = (str1 = '', str2 = '') => {
    const set1 = new Set(str1);
    const set2 = new Set(str2);
    let result = '';
    
    // Characters in str1 but not in str2
    for (let char of str1) {
        if (!set2.has(char)) {
            result += char;
        }
    }
    
    // Characters in str2 but not in str1
    for (let char of str2) {
        if (!set1.has(char)) {
            result += char;
        }
    }
    
    return result;
};

console.log(findUncommonWithSet("hello", "world"));
console.log(findUncommonWithSet("abc", "def"));

helwrd
abcdef

Comparison

Conclusion

Both approaches find uncommon characters between strings. The Set method is more efficient for larger strings, while the includes() method is simpler for basic use cases.