The challenge I am going to explain is the Jewels and Stones challenge, which, by the way, is a concise one. The description of the problem is as follows:

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

Thus, each jewel from the String J is compared with each stone I have. If the jewel matches the stone I have, it means that I have a jewel. After all the stones are compared, the function returns the number of jewels I have.

class Solution(object):
    def numJewelsInStones(self, J, S):
        """
        :type J: str
        :type S: str
        :rtype: int
        """
        
        '''
        J - types of stones that are jewels (J distinct)
        S - stones you have
        Each char in S is a type of stone I have 
        '''
        
        # take each stone from J an check if it is found in S
        return sum(stone in J for stone in S)

The following line:

return sum(stone in J for stone in S)

Can be restructured as follows:

count = 0

for jewel in J:
    for stone in S:
        if jewel == stone: 
            count += 1

return count

For the moment, I prefer the second solution because it is easier for a beginner like me to understand what is happening. Nonetheless, the first solution is more beautiful and quicker.

The statistics for the first code are: