Jewels and Stones

Problem Description

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.


Using Hashset to store all types of jewels from J. iterate stones S, check each stone whether equals to jewels in HashSet,

  • if current stone in set, then count++;

  • if current stone not in set, continue.

Complexity Analysis

Time Complexity: O(max(N, M))

Space Complexity: O(N)

  • N - length of string jewels J

  • M - length of string stones S


class Solution {
    public int numJewelsInStones(String J, String S) {
        if (J == null || S == null || S.length() == 0) return 0;
        int count = 0;
        Set<Character> set = new HashSet<>();
        for (char ch : J.toCharArray()) {
        for (char ch : S.toCharArray()) {
            if (set.contains(ch)) count++;
        return count;

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