Cracking the Code: LeetCode 389 (simple method)
Published: 2023-09-26 14:04:17


![leetcode 389](https://www.nileshblog.tech/wp-content/uploads/2023/09/leetcode-389.-find-the-difference-1536x864.png)
Have you ever felt the thrill of solving a complex puzzle, the satisfaction of finding the missing piece, or the joy of cracking a secret code? If you're a fan of challenges and enjoy problem-solving, LeetCode is the perfect platform for you. In this blog post, we're going to dive into the intriguing world of LeetCode, with a specific focus on problem 389 - "[Find The Difference] (https://www.nileshblog.tech/leetcode-389-find-the-difference/)." We'll unravel the secrets of this seemingly 'easy' problem, explore some strategies to tackle it, and provide you with a real-world example to solidify your understanding. So, are you ready to embark on this coding adventure with us?

## **The Mystery of LeetCode 389**

[LeetCode 389] (https://www.nileshblog.tech/leetcode-389-find-the-difference/), titled "Find The Difference," is categorized as an easy problem. But don't be fooled by its simplicity; it's a puzzle that can stump even the most experienced programmers. The problem statement goes like this: you are given two strings, s and t, which are both made up of lowercase English letters. The string t is generated by randomly shuffling string s and then adding one more letter at a random position. Your task is to find and return the letter that was added to t.

## **Cracking the Code**

To solve this puzzle, you can employ various strategies. One straightforward approach is to use dictionaries to count the frequency of each character in both strings and then compare them. The extra character in string t will be the one with a different count. Here's a step-by-step breakdown of this approach:
**Step 1: Create Dictionaries**
* Initialize two dictionaries, one for each string, to store the frequency of each character.
**Step 2: Count Frequencies**
* Iterate through each character in string s and update its count in the dictionary.
**Step 3: Subtract Frequencies**
* Iterate through each character in string t and update its count in the dictionary. If you encounter a character that's not in the dictionary, that's our missing character.
**Step 4: Find the Difference**
* Finally, compare the two dictionaries. The character with different counts is the answer.

## **Example: Solving LeetCode 389**

Let's illustrate the solution with an example. Suppose we have string s = "abcd" and string t = "abcde." The missing character in t is 'e.' Here's how our solution works:
1. We create two dictionaries: one for s and one for t.
2. We count the frequencies of characters in both strings.For s: {'a': 1, 'b': 1, 'c': 1, 'd': 1}For t: {'a': 1, 'b': 1, 'c': 1, 'd': 1, 'e': 1}
3. As we iterate through string t, we notice 'e' is not in the dictionary for s, which means 'e' is the missing character.
4. Our solution returns 'e' as the answer.

## **In Conclusion**

[LeetCode 389. Find The Difference] (https://www.nileshblog.tech/leetcode-389-find-the-difference/)," might be labeled as an easy problem, but it offers an excellent opportunity to sharpen your coding skills. By using dictionaries and counting character frequencies, you can confidently find the missing character and crack the code. So, if you're new to LeetCode or a seasoned coder looking for a quick challenge, give this problem a try. Happy coding!