3
\$\begingroup\$

I was trying to build a small utility function to check if an array is part of other array. It's testing if an array is a subset of another master array.

const masterArray = [1,2,3,4,5,6];
const candidateArray = [2,5,6];

//Test for subset.

//create a set from the two.
const s1 = new Set(masterArray.concat(candidateArray));

//Compare the sizes of the master array and the created set
//If the sizes are same, no new elements are added that means
//the candidate is complete subset.
s1.size === masterArray.length;

Can this be handled in a better way?

\$\endgroup\$
  • 2
    \$\begingroup\$ Wouldn't that fail if masterArray has duplicate elements? \$\endgroup\$ – Martin R Jul 16 '18 at 7:15
  • \$\begingroup\$ Good point. In my case the master array is always unique. \$\endgroup\$ – Sol Jul 16 '18 at 7:22
  • \$\begingroup\$ I can enhance it by comparing with a new set created from master array (new Set(masterArray)).size \$\endgroup\$ – Sol Jul 16 '18 at 8:46
  • \$\begingroup\$ I am looking for a way to do this with the type system (to check at compile time), if anyone knows how \$\endgroup\$ – Alexander Mills Jul 20 '19 at 22:48
1
\$\begingroup\$

So the key here is that I value code that's obvious to others.

Someone else is likely to try to stuff duplicate elements in unless it requires a type refactor; making const masterSet: Set<number> and using that throughout the code is a Good Idea. That "someone else" can often be you two months down the line, too.

In this case I would not say that the point of this length comparison is obvious; it will work but it takes some thinking -- that is why it has bugs with duplicate elements. In this case I would write something which I find more straightforward like,

function isSubsetOf<x>(sub: Iterable<x>, sup: Set<x>): boolean {
  for (const x of sub) {
    if (!sup.has(x)) {
      return false;
    }
  }
  return true;
}

I am also hinting through the type system how this function works by insisting that the subset is any iterable -- so I must be iterating through it.

With a restriction to an array and a set, one can get a little swankier by using the reduce function,

function isSubsetOf<x>(sub: Array<x>, sup: Set<x>): boolean {
  return sub.reduce((acc, x) => acc && sup.has(x), true);
}

but probably not all other readers will find that as intuitive as I do, and the former has slightly better runtime characteristics on very large lists that are obviously not subsets.

| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.