Scramble String — LeetCode 87 Python Solution

The problem

We can scramble a string s to get a string t using the following algorithm: If the length of the string is 1, stop. If the length of the string is > 1, do the following: Split the string into two non-empty substrings at a random index, i.e., if the string is s, divide it to x and y where s = x + y.

Example

Input

s1 = "great", s2 = "rgeat"

Output

true

Explanation

One possible scenario applied on s1 is:

Python solution

class Solution:
    def isScramble(self, s1: str, s2: str) -> bool:
        @cache
        def dfs(i: int, j: int, k: int) -> bool:
            if k == 1:
                return s1[i] == s2[j]
            for h in range(1, k):
                if dfs(i, j, h) and dfs(i + h, j + h, k - h):
                    return True
                if dfs(i + h, j, k - h) and dfs(i, j + k - h, h):
                    return True
            return False

        return dfs(0, 0, len(s1))

Complexity

Pattern: Dynamic Programming

Define a state, write the transition, and stop recomputing the same subproblem. LeetCode 87. Scramble String is filed here on both counts: the reference solution below belongs to the algorithm family this hub collects, and LeetCode tags it Dynamic Programming.

The dynamic programming guide has the Python template for the pattern and the 481 LeetCode problems that use it.

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