#2851

Hard

auto_awesomeState transition dynamic programming

LeetCode Problem Workspace

String Transformation

Find how many ways string s can be transformed into string t in exactly k operations using suffix rotations.

math string dynamic programming string matching

Problem Statement

You are given two strings, s and t, of equal length n, and an integer k. You can perform an operation where you select a suffix of string s and rotate it to the front, resulting in a new string. Your goal is to determine how many ways string s can be transformed into string t in exactly k operations.

Since the output can be large, return the answer modulo 10^9 + 7. Additionally, string t can only be constructed if it is a rotated version of string s, so make sure to verify if this is possible before proceeding with operations.

Examples

Example 1

Input: s = "abcd", t = "cdab", k = 2

Output: 2

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

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.

Solution Approach

State Transition Dynamic Programming

This problem can be solved using a state transition dynamic programming approach. Define a DP state where dp[i][j] represents the number of ways to reach string t from s in exactly i operations, with the jth suffix of s involved. Transition between states based on valid suffix selections and rotations, considering the constraints on k operations.

Cycle Detection and Rotation Matching

Since string t is a rotation of string s, we can detect possible rotations by checking for cyclic shifts of s. We need to account for all rotations and how they can be transformed into t, making sure the number of operations matches k. Efficiently track these rotations in the DP table to ensure accurate results.

Modulo Arithmetic Optimization

Given the large constraints on the input size, use modulo 10^9 + 7 to prevent overflow and ensure that calculations are within bounds. Each time a new state is computed in the DP table, apply the modulo operation to avoid exceeding the limit.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity depends on the final approach, but for a dynamic programming solution, it generally ranges from O(n * k) to O(n^2) due to state transitions. Space complexity also varies, typically depending on the number of states stored and the depth of the DP table, requiring O(n) or O(n * k) space.

What Interviewers Usually Probe

Candidate demonstrates understanding of state transition dynamic programming.
Candidate recognizes the importance of string rotations and modulo arithmetic in handling large numbers.
Candidate should explain how to optimize for large inputs and the trade-offs between time and space complexity.

Common Pitfalls or Variants

Common pitfalls

Forgetting to handle the modulo operation, leading to overflow or incorrect answers.
Misunderstanding the rotation property of string t and assuming it’s not derived from s.
Overcomplicating the DP state transitions, missing out on optimizing the table for efficient solutions.

Follow-up variants

What if k is much smaller than the number of operations required? Consider edge cases where k is limited.
How would you approach this problem if we allowed substring rotations instead of suffix rotations?
What if we need to count the transformations modulo some different number instead of 10^9 + 7?

FAQ

What is the main approach for solving the String Transformation problem?

The main approach is using state transition dynamic programming to track all possible transformations of string s to t with exactly k operations.

How do I ensure my solution handles large inputs in the String Transformation problem?

You must optimize the solution using modulo 10^9 + 7 to prevent overflow and ensure calculations stay within bounds, while also optimizing the DP table for large inputs.

What does it mean that string t can only be constructed if it's a rotated version of string s?

It means that string t must be a cyclic shift of s, and only in this case can you apply operations to transform s into t. This rotation property is crucial for validating the problem.

How do I handle large values of k efficiently?

By applying state transition dynamic programming and optimizing the space and time complexity of the DP table, you can handle very large values of k efficiently, especially with modulo arithmetic.

Can this problem be solved using a greedy approach?

A greedy approach is unlikely to work here due to the need for precise tracking of operations and state transitions. Dynamic programming is the most suitable approach.

terminal

Solution

Solution 1

#### Python3

100

"""
DP, Z-algorithm, Fast mod.
Approach
How to represent a string?
Each operation is just a rotation. Each result string can be represented by an integer from 0 to n - 1. Namely, it's just the new index of s[0].
How to find the integer(s) that can represent string t?
Create a new string s + t + t (length = 3 * n).
Use Z-algorithm (or KMP), for each n <= index < 2 * n, calculate the maximum prefix length that each substring starts from index can match, if the length >= n, then (index - n) is a valid integer representation.
How to get the result?
It's a very obvious DP.
If we use an integer to represent a string, we only need to consider the transition from zero to non-zero and from non-zero to zero. In other words, all the non-zero strings should have the same result.
So let dp[t][i = 0/1] be the number of ways to get the zero/nonzero string
after excatly t steps.
Then
dp[t][0] = dp[t - 1][1] * (n - 1).
All the non zero strings can make it.
dp[t][1] = dp[t - 1][0] + dp[t - 1] * (n - 2).
For a particular non zero string, all the other non zero strings and zero string can make it.
We have dp[0][0] = 1 and dp[0][1] = 0
Use matrix multiplication.
How to calculate dp[k][x = 0, 1] faster?
Use matrix multiplication
vector (dp[t - 1][0], dp[t - 1][1])
multiplies matrix
[0 1]
[n - 1 n - 2]
== vector (dp[t][0], dp[t - 1][1]).
So we just need to calculate the kth power of the matrix which can be done by fast power algorith.
Complexity
Time complexity:
O(n + logk)
Space complexity:
O(n)
"""


class Solution:
    M: int = 1000000007

    def add(self, x: int, y: int) -> int:
        x += y
        if x >= self.M:
            x -= self.M
        return x

    def mul(self, x: int, y: int) -> int:
        return int(x * y % self.M)

    def getZ(self, s: str) -> List[int]:
        n = len(s)
        z = [0] * n
        left = right = 0
        for i in range(1, n):
            if i <= right and z[i - left] <= right - i:
                z[i] = z[i - left]
            else:
                z_i = max(0, right - i + 1)
                while i + z_i < n and s[i + z_i] == s[z_i]:
                    z_i += 1
                z[i] = z_i
            if i + z[i] - 1 > right:
                left = i
                right = i + z[i] - 1
        return z

    def matrixMultiply(self, a: List[List[int]], b: List[List[int]]) -> List[List[int]]:
        m = len(a)
        n = len(a[0])
        p = len(b[0])
        r = [[0] * p for _ in range(m)]
        for i in range(m):
            for j in range(p):
                for k in range(n):
                    r[i][j] = self.add(r[i][j], self.mul(a[i][k], b[k][j]))
        return r

    def matrixPower(self, a: List[List[int]], y: int) -> List[List[int]]:
        n = len(a)
        r = [[0] * n for _ in range(n)]
        for i in range(n):
            r[i][i] = 1
        x = [a[i][:] for i in range(n)]
        while y > 0:
            if y & 1:
                r = self.matrixMultiply(r, x)
            x = self.matrixMultiply(x, x)
            y >>= 1
        return r

    def numberOfWays(self, s: str, t: str, k: int) -> int:
        n = len(s)
        dp = self.matrixPower([[0, 1], [n - 1, n - 2]], k)[0]
        s += t + t
        z = self.getZ(s)
        m = n + n
        result = 0
        for i in range(n, m):
            if z[i] >= n:
                result = self.add(result, dp[0] if i - n == 0 else dp[1])
        return result

Continue Practicing

#2719 count of integers #2999 count the number of powerful integers #3260 find the largest palindrome divisible by k