Leetcode Easy

## Problem

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.

## Solution

class Solution {
public int numJewelsInStones(String J, String S) {
HashSet<Character> jewels = new HashSet<>();

// Add all the elements to the hashSet
for (int i = 0; i < J.length(); i++) {
jewels.add(J.charAt(i));
}

// Count the
int numJewels = 0;
for (int i = 0; i < S.length(); i++) {
if (jewels.contains(S.charAt(i))) {
numJewels++;
}
}

return numJewels;
}
}


## How it works

First we loop through the string containing jewels and add them to a hashSet. This allows for us to loop through the second string and check if those stones are jewels in constant time, as the contains() method is O(1) for a hashSet.