Introduction

In this blog post, we will explore how to solve the LeetCode problem "459" 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

Given a string s, check if it can be constructed by taking a substring of it and appending multiple copies of the substring together. Example 1: Input: s = "abab" Output: true Explanation: It is the substring "ab" twice. Example 2: Input: s = "aba" Output: false Example 3: Input: s = "abcabcabcabc" Output: true Explanation: It is the substring "abc" four times or the substring "abcabc" twice. Constraints: 1 <= s.length <= 104 s consists of lowercase English letters.

Explanation

Here's the approach:

• Iterate through possible substring lengths: The core idea is to check all possible substring lengths from 1 up to half the length of the string. If a substring can form the entire string, its length must be a divisor of the string's length.

• Check for repetition: For each potential substring length, extract the substring and repeatedly concatenate it. Compare the result with the original string. If they match, the string can be formed by repeating the substring.

• Early exit: Return True as soon as a repeating substring is found. If no such substring is found after checking all possible lengths, return False.

• Runtime & Storage Complexity: O(n^2) time complexity, O(n) storage complexity.

Code

    def repeatedSubstringPattern(s: str) -> bool:
    """
    Checks if a string can be constructed by repeating a substring.

    Args:
        s: The input string.

    Returns:
        True if the string can be constructed by repeating a substring, False otherwise.
    """
    n = len(s)
    for i in range(1, n // 2 + 1):
        if n % i == 0:
            substring = s[:i]
            repeated_substring = substring * (n // i)
            if repeated_substring == s:
                return True
    return False