Algorithms Problem Solving: Jewels and Stones

This post is part of the Algorithms Problem Solving series. And it was originally published at TK's blog.

Problem description

This is the Jewels and Stones problem. The description looks like this:

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".

Examples

# input: J = "aA" | S = "aAAbbbb"
# output: 3

# input: J = "z", S = "ZZ"
# output: 0

Solution

At first, I tried to understand some corner cases. For example, what should I return if the J or the S is an empty string?

As my first solution, the brute force solution, I needed to look through the string. So I didn't need to care about the empty strings. For empty strings, it doesn't loop, it just returns the default counter, in this case, 0.

I just needed to loop through the J and for each character in the J string, I need to compare to each character of S. If they match, I increment the counter.

After looping through each character, just return the final counter.

def num_jewels_in_stones(J, S):
    jewels = 0

    for j in J:
        for s in S:
            if s == j:
                jewels += 1

    return jewels

print(num_jewels_in_stones("aA", "aAAbbbb")) # 3
print(num_jewels_in_stones("z", "ZZ")) # 0

This is a O(N^2) solution in terms of time complexity. Or more precisaly: O(len(J) * len(S)). For the space complexity, it is O(1) as we just store the value in a counter. If len(J) or len(S) increase, the used space keeps constant.

Just to iterate this solution, we can make it O(N) in terms of time complexity by using a hash table to store all characters as key and the counter as value.

def num_jewels_in_stones_opt(J, S):
    chars_counter = {}
    counter = 0

    for char in J:
        if char in chars_counter:
            chars_counter[char] += 1
        else:
            chars_counter[char] = 1

    for char in S:
        if char in chars_counter:
            counter += chars_counter[char]

    return counter

print(num_jewels_in_stones_opt("aA", "aAAbbbb")) # 3
print(num_jewels_in_stones_opt("z", "ZZ")) # 0

So, for each J's character. Verify if it is already in the chars_counter. If it is, just increment it. If not, initialize the counter.

Then, for each S's character, verify if this character is in the chars_counter. If it is, get it and add to the counter variable. If not, do nothing.

After this iteration, we need the final value of the counter. Just return it.

As we said before, it runs in O(N). Better than the first solution. But, for the worst-case scenario, the space complexity is O(N), worse than the first approach.

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Resources

• Learning Python: From Zero to Hero
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• Big-O Notation For Coding Interviews and Beyond
• Learn Python from Scratch
• Learn Object-Oriented Programming in Python
• Data Structures in Python: An Interview Refresher