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, 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
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