Javascript: Check if an Array is a Subset of Another Array

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There are many use cases where knowing if an array is a subset of another can be quite useful - and although usually immediately obvious to the human eye, it can be difficult to evaluate in code.

In the code below, arr2 is a subset of arr1, but arr3 is not.

let arr1 = [ 'a', 'b', 'c' ];
let arr2 = [ 'b', 'c' ];
let arr3 = [ 'c', 'd' ];

If we want to find out if an array is a subset of another array, the best way to evaluate this is by using the array every method. This method iterates through each element in an array and performs a test on it. If every element in the array passes that test, then the overall every method will return true.

For each evaluation, we have a "parent" array, and a "subset" array - where we want to check if the "subset" array is fully contained within the "parent" array.

To evaluate if one array is a subset of another, we can run every on the each element on the "subset" array. Within the every function, we can test if the "parent" array contains each element in the "subset" array. If it does, then every will return true. Otherwise, it'll return false.

This can be accomplished easily with a function like the one shown below - checkSubset will return true should the subsetArray be fully contained within the parentArray:

let arr1 = [ 'a', 'b', 'c' ];
let arr2 = [ 'b', 'c' ];
let arr3 = [ 'c', 'd' ];

let checkSubset = (parentArray, subsetArray) => {
    return subsetArray.every((el) => {
        return parentArray.includes(el)
    })
}

checkSubset(arr1, arr2); // returns true
checkSubset(arr1, arr3); // returns false

Oldest comments (3)

Revenity • Nov 19 '22 • Edited

You don't want to search at the beginning of the array every time you search an element in a subarray to make sure it exists in the array right?

function isSub<T = any>(parent: T[], sub: T[]) {
    let currentIndex = -1;
    for (const element of sub) 
        while (parent[++currentIndex] != element) 
            if (currentIndex >= parent.length)
                return false;
    return true;
}