# Solving Leetcode Interviews in Seconds with AI: First Unique Character in a String


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "387" 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
	> Given a string s, find the first non-repeating character in it and return its index. If it does not exist, return -1.   Example 1:  Input: s = "leetcode" Output: 0 Explanation: The character 'l' at index 0 is the first character that does not occur at any other index.  Example 2:  Input: s = "loveleetcode" Output: 2  Example 3:  Input: s = "aabb" Output: -1    Constraints:  1 <= s.length <= 105 s consists of only lowercase English letters.  

	# Explanation
	Here's the solution to find the first non-repeating character in a string, explained and implemented efficiently:

*   **Character Counting:** Use a hash map (dictionary in Python) to store the frequency of each character in the string.
*   **First Unique Character:** Iterate through the string again. For each character, check its frequency in the hash map. If the frequency is 1, return its index.
*   **Handle No Unique Character:** If the loop completes without finding a unique character, return -1.

*   **Runtime Complexity:** O(n), where n is the length of the string.
*   **Storage Complexity:** O(1), because the hash map will store at most 26 key-value pairs (lowercase English letters).

	
	# Code
	```python
	def firstUniqChar(s: str) -> int:
    """
    Finds the index of the first non-repeating character in a string.

    Args:
        s: The input string.

    Returns:
        The index of the first non-repeating character, or -1 if none exists.
    """

    char_counts = {}
    for char in s:
        char_counts[char] = char_counts.get(char, 0) + 1

    for i, char in enumerate(s):
        if char_counts[char] == 1:
            return i

    return -1
	```