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A few days ago I wrote a code that involved merging two large arrays without duplicates. My first reaction was to write a reduce function that initiated the accumulator, looped every element and checked whether it was already in the array:

const firstArr = new Array(200).fill(undefined).map((val, i) => `item${i}`)
const secondArr = new Array(250).fill(undefined).map((val, i) => `item${i}`)
const result = secondArr.reduce(
  (acc, item) => {
    return acc.includes(item) ? acc : [...acc, item]

Performance speed: 0.2749999985098839 milliseconds.

This solution worked well for me, but I wondered if I could improve the performance a little more, and ended up with a few solutions.

For testing purposes, each described method uses the same arrays:

const firstArr = new Array(200).fill(undefined).map((val, i) => `item${i}`)
const secondArr = new Array(250).fill(undefined).map((val, i) => `item${i}`)

also I will check the execution speed at the end of every function with

First solution

The traditional way to accomplish the task is to initialize an empty result array that contains the result items, create a loop for each array that has to be merged, and store the current element in the result array if it is not present. The presence is checked using indexOf that will return -1 if not.

const result = []
for (let i = 0; i < firstArr.length; i++) {
  if (result.indexOf(firstArr[i]) == -1) result.push(firstArr[i])
for (let i = 0; i < secondArr.length; i++) {
  if (result.indexOf(secondArr[i]) == -1) result.push(secondArr[i])

Performance speed: 0.41500001680105925 milliseconds.

Second solution

Using ES5 we can merge the two input arrays without removing the duplicates with .concat() and then loop again through the result array and remove duplicates using indexOf.

const concatArr = firstArr.concat(secondArr)
const result = concatArr.filter((item, idx) => concatArr.indexOf(item) === idx)

Performance speed: 0.5300000193528831 milliseconds.

Third solution

Overall, the reduce method at the beginning still has the better performance. Between the last two solutions there is no significant difference in complexity and execution speed, but the second is cleaner and easier to read. We can go a step further and take into account the ES6 Set feature, which takes all values from an iterable object, in our case an array, but since every element in a set must be unique in the collection, it skips the duplicates without any effort for us.

When creating a Set keep in mind that null is treated like undefined and that it is a different concept from an array: a Set is a “keyed collection” while the second is an “indexed collection”

Here is the code:

const result = [ Set([...firstArr, ...secondArr])]

Performance speed: 0.13499998021870852 milliseconds.

Since we don’t want the result to be a new Set we will just destructure it into an array.


The last solution is what I was looking for, it is the most readable and it has the lowest complexity, which brings a great performance compared to the other two. I ended up bringing that piece of code into my application abstracting it into a new function. Thanks, I hope this was helpful enough 😉