Finding deviations in two Number arrays in JavaScript

We are required to write a JavaScript function that takes in two number arrays and returns the elements that are not common to both arrays (symmetric difference).

For example, if the two arrays are:

const arr1 = [2, 4, 2, 4, 6, 4, 3];
const arr2 = [4, 2, 5, 12, 4, 1, 3, 34];

Then the output should be:

[6, 5, 12, 1, 34]

Method 1: Using indexOf() Method

This approach uses nested loops to check if elements exist in the other array:

const arr1 = [2, 4, 2, 4, 6, 4, 3];
const arr2 = [4, 2, 5, 12, 4, 1, 3, 34];

const deviations = (first, second) => {
    const res = [];
    
    // Check elements in first array that don't exist in second
    for(let i = 0; i < first.length; i++){
        if(second.indexOf(first[i]) === -1){
            res.push(first[i]);
        }
    }
    
    // Check elements in second array that don't exist in first
    for(let j = 0; j < second.length; j++){
        if(first.indexOf(second[j]) === -1){
            res.push(second[j]);
        }
    }
    
    return res;
};

console.log(deviations(arr1, arr2));

[6, 5, 12, 1, 34]

Method 2: Using Set for Better Performance

Using Set provides O(1) lookup time instead of O(n) with indexOf:

const arr1 = [2, 4, 2, 4, 6, 4, 3];
const arr2 = [4, 2, 5, 12, 4, 1, 3, 34];

const findDeviations = (first, second) => {
    const set1 = new Set(first);
    const set2 = new Set(second);
    const result = [];
    
    // Add elements from first array not in second
    for(let item of set1) {
        if(!set2.has(item)) {
            result.push(item);
        }
    }
    
    // Add elements from second array not in first
    for(let item of set2) {
        if(!set1.has(item)) {
            result.push(item);
        }
    }
    
    return result;
};

console.log(findDeviations(arr1, arr2));

[6, 5, 12, 1, 34]

Method 3: Using filter() Method

A more functional approach using array methods:

const arr1 = [2, 4, 2, 4, 6, 4, 3];
const arr2 = [4, 2, 5, 12, 4, 1, 3, 34];

const getDeviations = (first, second) => {
    const unique1 = [...new Set(first)];
    const unique2 = [...new Set(second)];
    
    const diff1 = unique1.filter(x => !unique2.includes(x));
    const diff2 = unique2.filter(x => !unique1.includes(x));
    
    return [...diff1, ...diff2];
};

console.log(getDeviations(arr1, arr2));

[6, 5, 12, 1, 34]

Performance Comparison

Conclusion

The Set-based approach offers the best performance for large arrays, while indexOf() is simpler for smaller datasets. Choose the method based on your data size and performance requirements.