Leetcode Easy


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


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


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++) {

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

        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.