Leetcode #2851: String Transformation
In this guide, we solve Leetcode #2851 String Transformation in Python and focus on the core idea that makes the solution efficient.
You will see the intuition, the step-by-step method, and a clean Python implementation you can use in interviews.
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.
Quick Facts
- Difficulty: Hard
- Premium: No
- Tags: Math, String, Dynamic Programming, String Matching
Intuition
The problem breaks into overlapping subproblems, so caching results prevents exponential repetition.
A carefully chosen DP state captures exactly what we need to build the final answer.
Approach
Define the DP state and recurrence, then compute states in the correct order.
Optionally compress space once the recurrence is clear.
Steps:
- Choose a DP state definition.
- Write the recurrence and base cases.
- Compute states in the correct order.
Example
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".
Python Solution
"""
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
Complexity
The time complexity is O(n·m) (typical). The space complexity is O(n·m) or optimized.
Edge Cases and Pitfalls
Watch for boundary values, empty inputs, and duplicate values where applicable. If the problem involves ordering or constraints, confirm the invariant is preserved at every step.
Summary
This Python solution focuses on the essential structure of the problem and keeps the implementation interview-friendly while meeting the constraints.