Best way to remove elements from an array1 which are present in array2 Javascript?

Question

Let's say we have the following JavaScript array

var array1 = ['Teams' , 'Chat' , 'Zoom' , 'Slack' ] 

var array2 = [ 'Zoom' , 'Chat']

How to make a new array in which only those elements of array1 present which are not in array2?

New Array should look like

new_array = [ 'Teams' , 'Slack' ]

this helps? https://stackoverflow.com/questions/1187518/how-to-get-the-difference-between-two-arrays-in-javascript — cmgchess, Mar 24 '22 at 19:59

PeterKA · Answer 1 · 2022-03-24T20:06:38.023

2

Use Array#filter() method:

const newArray = array1.filter(e => !array2.includes(e));

DEMO

const array1 = ['Teams' , 'Chat' , 'Zoom' , 'Slack' ];
const array2 = [ 'Zoom' , 'Chat'];

const newArray = array1.filter(e => !array2.includes(e));

console.log( newArray );

edited Mar 24 '22 at 20:06

answered Mar 24 '22 at 20:02

PeterKA

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Maybe use `const` for the arrays as well. – Andy Mar 24 '22 at 20:05

Mushroomator · Answer 2 · 2022-03-24T20:23:24.843

This will run in O(n) time in best, worst and average case for n elements in array1 as the lookup in the Set happens in O(1) and will return the correct result.

Using Array#includes() will result in a worst case runtime of O(n * m) for n elements in array1 and m elements in array2 when all elements in array1 and array2 are different. And even if they are not you would on average need m / 2 steps to find a match in array2 using Array#includes(), which would still result in an average case runtime of O(n * m).

const array1 = ['Teams' , 'Chat' , 'Zoom' , 'Slack' ] 

const array2 = [ 'Zoom' , 'Chat']

// Create a Set for lookups in O(1); Creation will take O(n).
const array2Set = new Set(array2);
// Only return elements that are not in the Set in O(n) 
const newArray = array1.filter(item => !array2Set.has(item))

// print result
console.log(newArray);

Best way to remove elements from an array1 which are present in array2 Javascript?

2 Answers2