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# What’s your alternative solution? Challenge #27

This is series of daily JavaScript coding challenges... for both beginners and advanced users.

Each day I’m gone present you a very simple coding challenge, together with the solution. The solution is intentionally written in a didactic way using classic JavaScript syntax in order to be accessible to coders of all levels.

Solutions are designed with increase level of complexity.

## Today’s coding challenge

``````Create a function that will receive two arrays of numbers as arguments and return an array composed of all the numbers that are either in the first array or second array but not in both

``````

(scroll down for solution)

## Code newbies

If you are a code newbie, try to work on the solution on your own. After you finish it, or if you need help, please consult the provided solution.

You can solve it using functional concepts or solve it using a different algorithm... or just solve it using the latest ES innovations.

By providing a new solution you can show code newbies different ways to solve the same problem.

## Solution

``````// Solution for challenge25

var ar1 = [1, 2, 3, 10, 5, 3, 14];
var ar2 = [1, 4, 5, 6, 14];

var ar = mergeExclusive(ar1, ar2);
println(ar);

function mergeExclusive(ar1, ar2)
{
var ar = [];

for(let el of ar1)
{
if (!ar2.includes(el))
{
ar.push(el);
}
}

for(let el of ar2)
{
if (!ar1.includes(el))
{
ar.push(el);
}
}

return ar;
}

``````

To quickly verify this solution, copy the code above in this coding editor and press "Run".

Note: The solution was originally designed for codeguppy.com environment, and therefore is making use of `println`. This is the almost equivalent of `console.log` in other environments. Please feel free to use your preferred coding playground / environment when implementing your solution.

## Discussion

Curtis Fenner

The given algorithm is simple, but runs in quadratic time. You can use `Set`s to make the `includes` check faster.

``````const inLeft = new Set(left);
const inRight = new Set(right);
return left.filter(x => !inRight.has(x)).concat(right.filter(x => !inLeft has(x));
``````