Fastest way to check if array contains 2 different values?

Question

Consider the following arrays:

['a', 'b', 'a'] //method should return true
['a', 'b', 'c'] //method should return true
['a', 'c', 'c'] //method should return false

I want to write a method that most efficiently checks to see if both 'a' and 'b' exist in the array. I know I can do this in a simple for loop

let a_counter = 0;
let b_counter = 0;
for (let i = 0; i < array.length; i++) {
 if (array[i] === 'a') {
 a_counter++;
 }
 if (array[i] === 'b') {
 b_counter++;
 }
}
return (a_counter > 0 && b_counter > 0);

But this isn't very short. I can do indexOf but that will loop through twice. I have also considered using a set as below:

const letter_set = new Set(array)
return (letter_set.has('a') && letter_set.has('b')) 

But I am pretty unfamiliar with sets and don't know if this solution could potentially be more expensive than just looping. I know that has() operations should be faster than array iterations but constructing the set probably takes at least O(N) time (I'm assuming).

Is there a clean and efficient way to find multiple elements in an array? ES6 answers welcome

• chevybow: Can the person who downvoted explain why? :)

Answer 1

You can use every and includes to do this check.

So we are saying every item must be included in the array.

function contains(arr, ...items) {
 return items.every(i => arr.includes(i))
}
console.log(contains(['a', 'b', 'a'], 'a', 'b'))
console.log(contains(['a', 'c', 'c'], 'a', 'b'))
console.log(contains(['a', 'b', 'c'], 'a', 'b', 'c'))
console.log(contains(['a', 'b', 'c', 'd'], 'a', 'b', 'c', 'd', 'e'))

• chevybow: I really like the readability of this answer. For some reason I didn't consider using every()

Answer 2

You could use just the Set and check if the wanted items are in the items array.

const
 check = (items, wanted) => wanted.every(Set.prototype.has, new Set(items));
console.log(check(['a', 'b', 'a'], ['a', 'b'])); // true
console.log(check(['a', 'b', 'c'], ['a', 'b'])); // true
console.log(check(['a', 'c', 'c'], ['a', 'b'])); // false

Answer 3

array.includes('a') && array.includes('b')

includes seems like a real handy way to check for specific elements, even if there is more than one.

• chevybow: @NinaScholz I made an edit for another test case. I don't mind if there's other elements (at least for the current function I'm working on)

Answer 4

Not as compact as the other examples, but it does do the job in single run.

const arr1 = ['a', 'b', 'a']; //method should return true
const arr2 = ['a', 'c', 'c']; //method should return false
const arr3 = ['a', 'b', 'c']; //method should return true
const reducer = ({ a, b }, char) => ({
 a: a || char === 'a',
 b: b || char === 'b'
});
const includesAnB = arr => {
 const { a, b } = arr.reduce(reducer, {});
 return a && b;
}
console.log(includesAnB(arr1));
console.log(includesAnB(arr2));
console.log(includesAnB(arr3));