Introduction

In this blog post, we will explore how to solve the LeetCode problem "466" 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

We define str = [s, n] as the string str which consists of the string s concatenated n times. For example, str == ["abc", 3] =="abcabcabc". We define that string s1 can be obtained from string s2 if we can remove some characters from s2 such that it becomes s1. For example, s1 = "abc" can be obtained from s2 = "abdbec" based on our definition by removing the bolded underlined characters. You are given two strings s1 and s2 and two integers n1 and n2. You have the two strings str1 = [s1, n1] and str2 = [s2, n2]. Return the maximum integer m such that str = [str2, m] can be obtained from str1. Example 1: Input: s1 = "acb", n1 = 4, s2 = "ab", n2 = 2 Output: 2 Example 2: Input: s1 = "acb", n1 = 1, s2 = "acb", n2 = 1 Output: 1 Constraints: 1 <= s1.length, s2.length <= 100 s1 and s2 consist of lowercase English letters. 1 <= n1, n2 <= 106

Explanation

Here's the breakdown of the solution:

• Pattern Detection: The core idea is to recognize repeating patterns in how s2 is formed within repeated concatenations of s1. We iterate through the repetitions of s1 and track how much of s2 we've covered. After some iterations, a repeating pattern should emerge.

• Memoization: We use memoization to store the state of the s2 index at the start of each s1 repetition. This helps us detect cycles and avoid redundant computations. Specifically, we store s2_index when we start each s1 repetition.

• Cycle Exploitation: Once a cycle is detected, we calculate how many complete cycles of s1 are present in the remaining repetitions of s1 and, accordingly, how many more s2 strings can be formed from these cycles.

• Runtime & Space Complexity: O(len(s1) * len(s2)) runtime, O(len(s1) * len(s2)) storage

Code

    def getMaxRepetitions(s1: str, n1: int, s2: str, n2: int) -> int:
    """
    Calculates the maximum integer m such that str = [str2, m] can be obtained from str1.

    Args:
        s1: The first string.
        n1: The number of times to repeat s1.
        s2: The second string.
        n2: The number of times to repeat s2.

    Returns:
        The maximum integer m.
    """

    s1_len, s2_len = len(s1), len(s2)
    s1_count, s2_count = 0, 0
    s2_index = 0
    memo = {}

    for i in range(n1):
        if s2_index in memo:
            prev_s1_count, prev_s2_count = memo[s2_index]
            remaining_s1 = n1 - i
            cycle_len = i - prev_s1_count
            cycle_s2 = s2_count - prev_s2_count

            num_cycles = remaining_s1 // cycle_len
            s2_count += cycle_s2 * num_cycles
            i += cycle_len * num_cycles

        else:
            memo[s2_index] = (i, s2_count)

        for j in range(s1_len):
            if s1[j] == s2[s2_index]:
                s2_index += 1
                if s2_index == s2_len:
                    s2_count += 1
                    s2_index = 0

    return s2_count // n2