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.
[code lang="java"]
class Solution {
    public int numJewelsInStones(String J, String S) {
        
    }
}
[code]

Idea – 1

While processing letter c in S, we need to answer “Is c a jewel?” To answer efficiently we use hash set (array based) for all the jewels. Time complexity is O( max(|J|, |S|) ) and space complexity is O(1). 
[code lang="java"]
class Solution {
    public int numJewelsInStones(String J, String S) {
        boolean[] set = new boolean[128];
        for(int i = 0; i < J.length(); ++i)
        {
            int j = J.charAt(i);
            set[j] = true;
        }
        
        int jewelCount = 0;
        for(int i = 0; i < S.length(); ++i)
        {
            int j = S.charAt(i);
            if(set[j])
            {
                ++jewelCount;
            }
        }
        
        return jewelCount;
    }
}
[code]

Runtime: 1 ms, faster than 98.37% of Java online submissions for Jewels and Stones.Memory Usage: 36.8 MB, less than 87.95% of Java online submissions for Jewels and Stones.

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