Introduction

In this blog post, we will explore how to solve the LeetCode problem "686" 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 two strings a and b, return the minimum number of times you should repeat string a so that string b is a substring of it. If it is impossible for b​​​​​​ to be a substring of a after repeating it, return -1. Notice: string "abc" repeated 0 times is "", repeated 1 time is "abc" and repeated 2 times is "abcabc". Example 1: Input: a = "abcd", b = "cdabcdab" Output: 3 Explanation: We return 3 because by repeating a three times "abcdabcdabcd", b is a substring of it. Example 2: Input: a = "a", b = "aa" Output: 2 Constraints: 1 <= a.length, b.length <= 104 a and b consist of lowercase English letters.

Explanation

Here's the breakdown of the solution:

• Determine the minimum repetitions: Start by repeating string a a sufficient number of times to ensure the repeated string's length is at least the length of b.
• Substring Check: Check if b is a substring of the repeated a. If found, return the number of repetitions.

• One More Repetition Check: If b is not found, repeat a one more time and check again. This handles cases where b spans across the boundary of the initial repetitions. If still not found return -1.

• Runtime Complexity: O(M * N) where M is the number of repetitions and N is the length of string b.

• Space Complexity: O(M * K) where M is the number of repetitions and K is the length of string a.

Code

    def repeatedStringMatch(a: str, b: str) -> int:
    """
    Given two strings a and b, return the minimum number of times you should repeat string a
    so that string b is a substring of it. If it is impossible for b​​​​​​ to be a substring of a after
    repeating it, return -1.
    """
    n = len(a)
    m = len(b)
    repetitions = (m + n - 1) // n  # Minimum repetitions needed

    repeated_a = a * repetitions
    if b in repeated_a:
        return repetitions

    repeated_a += a
    if b in repeated_a:
        return repetitions + 1

    return -1