# Solving Leetcode Interviews in Seconds with AI: Count The Repetitions


	# 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](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
	> 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
	```python
	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
	```