Introduction

In this blog post, we will explore how to solve the LeetCode problem "771" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

You're given strings jewels representing the types of stones that are jewels, and stones representing the stones you have. Each character in stones is a type of stone you have. You want to know how many of the stones you have are also jewels. Letters are case sensitive, so "a" is considered a different type of stone from "A". Example 1: Input: jewels = "aA", stones = "aAAbbbb" Output: 3 Example 2: Input: jewels = "z", stones = "ZZ" Output: 0 Constraints: 1 <= jewels.length, stones.length <= 50 jewels and stones consist of only English letters. All the characters of jewels are unique.

Explanation

Here's the solution to the "Jewels and Stones" problem:

• High-Level Approach:

  • Create a set of the jewels string to allow for O(1) lookups.
  • Iterate through the stones string and check if each stone is present in the jewels set.
  • Increment a counter for each stone that is a jewel.

• Complexity Analysis:

  • Runtime Complexity: O(J + S), where J is the length of the jewels string and S is the length of the stones string.
  • Storage Complexity: O(J), for storing the set of jewels.

Code

    def numJewelsInStones(jewels: str, stones: str) -> int:
    """
    Counts the number of stones that are also jewels.

    Args:
        jewels: A string representing the types of stones that are jewels.
        stones: A string representing the stones you have.

    Returns:
        The number of stones you have that are also jewels.
    """

    jewel_set = set(jewels)
    count = 0
    for stone in stones:
        if stone in jewel_set:
            count += 1
    return count