Lookup an array against another with performance

Question

I got an initial array, I am checking against another array to find how many objects have at least one instance of the Domain in data.

I wonder if there is a more performant way to achieve the same goal.

const data = [
 {
 Domain: 'google.com',
 '# Reocurring Domains': 0
 },
 {
 Domain: 'apple.com',
 '# Reocurring Domains': 0
 },
 {
 Domain: 'facebook.com',
 '# Reocurring Domains': 0
 }
]
const domains = [
 {
 'google.com': true,
 'microsoft.com': true,
 'google.com': true
 },
 {
 'apple.com': true,
 'microsoft.com': true,
 'twitter.com': true
 },
 {
 'facebook.com': true,
 'apple.com': true,
 'facebook.com': true
 }
]
for (const obj of data) {
 let count = 1
 for (const entry of domains) {
 if (entry[obj.Domain]) {
 obj['# Reocurring Domains'] = count++
 }
 }
}
console.log(data)

In there any way to this with a more performant approach?

Answer 1

You could create a look up table (a simple object) where you can pre calculate the amount of times a domain is in the repeated in the domains array, this reduces the algorithmic complexity of your current solution which is n * m (dataElements * domainElements)

Whereas my suggestion has a lower complexity of (n + m)

const data = [
 {
 Domain: 'google.com',
 '# Reocurring Domains': 0
 },
 {
 Domain: 'apple.com',
 '# Reocurring Domains': 0
 },
 {
 Domain: 'facebook.com',
 '# Reocurring Domains': 0
 }
]
const domains = [
 {
 'google.com': true,
 'microsoft.com': true,
 'google.com': true
 },
 {
 'apple.com': true,
 'microsoft.com': true,
 'twitter.com': true
 },
 {
 'facebook.com': true,
 'apple.com': true,
 'facebook.com': true
 }
]
const domainsObj = {}
domains.forEach(domain => {
 Object.keys(domain).forEach(key => key in domainsObj ? domainsObj[key] += 1: domainsObj[key] = 1)
})
data.forEach(element => element['# Reocurring Domains'] = domainsObj[element.Domain]);
console.log(data);

This approach increases memory usage, but improves calculation speed.

• Álvaro: Thanks Daniel, when I check in jsbench, it says otherwise jsbench.me/7lkl9a1bf5/2
• Daniel Rodríguez Meza: Well for a small amount of data it could be the case that this approach is not that good, but if we consider the case that your data is going to be large enough, this solution should be better :)