# Solving Leetcode Interviews in Seconds with AI: Jewels and Stones


	# 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](https://www.chatmagic.app), 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
	```python
	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
	```