Intersection of two arrays II JavaScript

Urfan Guliyev

Posted on Jul 12, 2020 • Updated on Jul 19, 2020

Today I am going to show how to solve the Leetcode Intersection of Two Arrays algorithm problem.

Here is the problem:

Solution:
There are no built-in methods in JavaScript for computing an intersection. That's why I use a new data structure, Set, which was introduced in ECMAScript 2015. As you know, set objects are collections of unique values.

First, I use Set to store both arrays. Then, I create an array from a set1 using the spread operator. I filter through the array to select the elements that are also in the second set object (set2).

var intersection = function(nums1, nums2) { let set1 = new Set(nums1); let set2 = new Set(nums2); let output = [...set1].filter(x => set2.has(x)) return output };

Stef van Hooijdonk - Oct 26

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Given two arrays, write a function to compute their intersection.

Example 1:

Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2]

Example 2:

Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [4,9]

Note:

Each element in the result should appear as many times as it shows in both arrays.
The result can be in any order.

Follow up:

What if the given array is already sorted? How would you optimize your algorithm?
What if nums1's size is small compared to nums2's size? Which algorithm is better?
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once?

/** * @param {number[]} nums1 * @param {number[]} nums2 * @return {number[]} */ const intersect = (nums1, nums2) => { const map = nums1.reduce( (acc, num1) => { if (acc[num1]) { acc[num1] += 1 } else { acc[num1] = 1 } return acc }, {}, ) return nums2.filter((num2) => { if (map[num2]) { map[num2] -= 1 return true } return false }) }

To get the difference of two arrays in JavaScript, you can use the combination of filter and includes. Imagine you have the following two arrays and you want to find out which elements are included in both:

const veggies = ['🍅', '🥕', '🥒']; const fruits = ['🍓', '🍌', '🍅'];

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To get their intersection, you can use the following call:

const intersection = veggies.filter(v => fruits.includes(v));

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To get their differencem eg.: only the ones that veggies include, you can do the following:

const difference = veggies.filter(v => !fruits.includes(v)); const difference = fruits.filter(f => !veggies.includes(f));

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And in order to get the symmetric difference of two arrays — meaning leave out the ones that are in both arrays — you need to filter both of them and either use concat or destructuring:

const symmetricDifference = [ ...veggies.filter(v => !fruits.includes(v)), ...fruits.filter(f => !veggies.includes(f)) ];

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Looking to improve your skills? Check out our interactive course to master JavaScript from start to finish.

If you need to rely on the same logic multiple times in your codebase, you can reuse them with the following function calls:

const intersect = (a, b) => a.filter(a => b.includes(a)); const difference = (a, b) => a.filter(a => !b.includes(a)); const symmetric = (a, b) => [ ...a.filter(a => !b.includes(a)), ...b.filter(b => !a.includes(b)) ];