# Solving Leetcode Interviews in Seconds with AI: Consecutive Characters


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1446" 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
	> The power of the string is the maximum length of a non-empty substring that contains only one unique character. Given a string s, return the power of s.   Example 1:  Input: s = "leetcode" Output: 2 Explanation: The substring "ee" is of length 2 with the character 'e' only.  Example 2:  Input: s = "abbcccddddeeeeedcba" Output: 5 Explanation: The substring "eeeee" is of length 5 with the character 'e' only.    Constraints:  1 <= s.length <= 500 s consists of only lowercase English letters.  

	# Explanation
	Here's a solution to find the power of a string, focusing on efficiency:

*   **Iterate and Track:** Iterate through the string, keeping track of the current character and the length of the current substring of identical characters.
*   **Update Maximum:** Update the maximum power (length) found so far whenever the current substring ends or at the end of the string.
*   **Reset Counter:** Reset the current substring length when a different character is encountered.

*   **Runtime Complexity:** O(n), where n is the length of the string. **Storage Complexity:** O(1).

	
	# Code
	```python
	def max_power(s: str) -> int:
    """
    Calculates the power of a string.

    The power of the string is the maximum length of a non-empty substring
    that contains only one unique character.

    Args:
        s: The input string.

    Returns:
        The power of the string.
    """
    max_len = 0
    curr_len = 0
    curr_char = ''

    for char in s:
        if char == curr_char:
            curr_len += 1
        else:
            max_len = max(max_len, curr_len)
            curr_char = char
            curr_len = 1

    max_len = max(max_len, curr_len)  # Check the last substring
    return max_len
	```