Items also occurs in a set

Question

The task

You're given strings J representing the types of stones that are jewels, and S representing the stones you have. Each character in S is a type of stone you have. You want to know how many of the stones you have are also jewels.

The letters in J are guaranteed distinct, and all characters in J and S are letters. Letters are case sensitive, so "a" is considered a different type of stone from "A".

Example 1:

Input: J = "aA", S = "aAAbbbb" Output: 3

Example 2:

Input: J = "z", S = "ZZ" Output: 0

Note:

S and J will consist of letters and have length at most 50. The characters in J are distinct.

My functional solution

const findNumberOfJewels = (j,s) =>  [...s].reduce((num, x) => [...j].includes(x) ? num + 1 : num , 0);

console.log(findNumberOfJewels("aA", "aAAbbbb"));

My imperative solution

function findNumberOfJewels2(j,s) {
  let res = 0;
  const set = new Set(j);
  for (const x of s) {
    if (set.has(x)) { ++res; }
  }
  return res;
};

console.log(findNumberOfJewels2("aA", "aAAbbbb"));

Answer 1

The functional approach is almost certainly faster despite having worse big-O — \$O(j \cdot s)\$ functional vs \$O(j + s)\$ imperative — because linear searches of small arrays are very fast.

You don't need to destructure j and adding a boolean to a number coerces the boolean to 0 or 1.

sum is a better name than num and c is a good name for a character iterator.

const findNumberOfJewels = (j,s) => [...s].reduce( (sum, c) => sum + j.includes(c),  0 );