# LeetCode Challenge #3: Jewels and Stones

### Problem Statement
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```

### Step by Step Solution
**Identify patterns** After reading the problem and examples, we can conclude 3 things:
1. The output of this problem is an integer that represents how many characters of String J exist in String S.
2. We have to compare each character (char) of J to each character (char) of S. 
3. If character of J is found in S, we increase the output count by 1.

**Write the Pseudocode** The following solution can be written as:
1. Initialize the output integer count to 0.
2. Have a for loop that iterates through String S.
3. Inside that for loop, have a for loop that iterates through String J. If char at S[i] == char at J[j], increase count by 1. Else, move onto the next index.
4. Finally, after the outer for loop finishes, return the count output.

>Remember: We want to check if a char of J is found in each char of S. So for example, when the outer for loop is at S[0], we will have the inner for loop iterate through J[0] to J[length-1] to check if S[0] == J[j]. If there is, increase the count variable by 1. Else, do nothing. Then, the outer for loop will move onto S[1] and the inner for loop will again check if J[0] to J[length-1] matches S[1] and so on. The loop stops after it reaches the final character of S.

**Let's get coding!** Now, let's try to put the pseudocode into Java:
```
class Solution {
    public int numJewelsInStones(String J, String S) {
        
        //Initialize count
        int count = 0;
        
        for(int i=0; i< S.length(); i++){
            for(int j=0; j<J.length(); j++){
                if(S.charAt(i) == J.charAt(j)){ //check if there's a match
                    count++;
                    break;
                }
            }
        }
        return count;
    }
}
```
### Conclusion
This problem may seem difficult to understand for beginners but once you have a clear understanding about strings and for loops, it is definitely not a hard problem. I hope I explained it clear enough but feel free to ask any clarifying questions on my profile [here](https://hashnode.com/@victoria). Keep practicing and good luck to anyone who's challenging LeetCode too!