# Solving Leetcode Interviews in Seconds with AI: String Transformation


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "2851" 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
	> You are given two strings s and t of equal length n. You can perform the following operation on the string s:  Remove a suffix of s of length l where 0 < l < n and append it at the start of s. 	For example, let s = 'abcd' then in one operation you can remove the suffix 'cd' and append it in front of s making s = 'cdab'.  You are also given an integer k. Return the number of ways in which s can be transformed into t in exactly k operations. Since the answer can be large, return it modulo 109 + 7.   Example 1:  Input: s = "abcd", t = "cdab", k = 2 Output: 2 Explanation:  First way: In first operation, choose suffix from index = 3, so resulting s = "dabc". In second operation, choose suffix from index = 3, so resulting s = "cdab".  Second way: In first operation, choose suffix from index = 1, so resulting s = "bcda". In second operation, choose suffix from index = 1, so resulting s = "cdab".  Example 2:  Input: s = "ababab", t = "ababab", k = 1 Output: 2 Explanation:  First way: Choose suffix from index = 2, so resulting s = "ababab".  Second way: Choose suffix from index = 4, so resulting s = "ababab".    Constraints:  2 <= s.length <= 5 * 105 1 <= k <= 1015 s.length == t.length s and t consist of only lowercase English alphabets.  

	# Explanation
	Here's a solution to the problem, focusing on efficiency and clarity:

*   **Key Idea:** The core idea is to recognize that each operation is essentially a right cyclic shift of the string. We need to find the number of sequences of `k` cyclic shifts that transform `s` into `t`. We can use dynamic programming or matrix exponentiation. Since `k` can be large, matrix exponentiation is more efficient.

*   **Transition Matrix:** We build a transition matrix where `matrix[i][j]` represents the number of ways to go from a shift of `i` to a shift of `j` in one operation.  This depends only on the length of the string `n`.

*   **Matrix Exponentiation:** Raise the transition matrix to the power of `k`. This gives us the number of ways to go from any shift to any other shift in `k` operations.

*   **Runtime & Storage Complexity:** O(n<sup>3</sup> log k) time, O(n<sup>2</sup>) space, where n is the length of the string and k is the number of operations.

	
	# Code
	```python
	def solve():
    s = input()
    t = input()
    k = int(input())
    n = len(s)
    MOD = 10**9 + 7

    def multiply(A, B):
        C = [[0] * n for _ in range(n)]
        for i in range(n):
            for j in range(n):
                for l in range(n):
                    C[i][j] = (C[i][j] + A[i][l] * B[l][j]) % MOD
        return C

    def power(A, k):
        result = [[0] * n for _ in range(n)]
        for i in range(n):
            result[i][i] = 1

        while k > 0:
            if k % 2 == 1:
                result = multiply(result, A)
            A = multiply(A, A)
            k //= 2
        return result

    # Find the initial shift of s
    initial_shift = 0
    for i in range(n):
        if s == t:
            break
        s = s[-1] + s[:-1]
        initial_shift += 1
    else:
        if s != t:
            print(0)
            return

    # Build the transition matrix
    transition_matrix = [[0] * n for _ in range(n)]
    for i in range(n):
        for j in range(1, n):
            transition_matrix[i][(i + j) % n] = 1

    # Raise the transition matrix to the power of k
    final_matrix = power(transition_matrix, k)

    # The answer is the number of ways to go from initial_shift to 0
    print(final_matrix[initial_shift % n][0])


solve()
	```