Jewels and stones

Time: O(M+N); Space: O(N); 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

Constraints:

  • S and J will consist of letters and have length at most 50.

  • The characters in J are distinct.

Hints:

  1. For each stone, check if it is a jewel.

[1]:

class Solution1(object):
    """
    Time: O(M+N)
    Space: O(N)
    """
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        lookup = set(J)

        return sum(s in lookup for s in S)

[2]:

s = Solution1()

J = "aA"
S = "aAAbbbb"
assert s.numJewelsInStones(J, S) == 3

J = "z"
S = "ZZ"
assert s.numJewelsInStones(J, S) == 0