#87

Hard

auto_awesomeState transition dynamic programming

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Scramble String

Scramble String is a dynamic programming problem where we determine if one string can be scrambled to form another using recursive state transitions.

string dynamic programming

Problem Statement

You are given two strings, s1 and s2, both of the same length. The task is to determine if s2 is a scrambled version of s1. A string is considered scrambled if it can be transformed into another by recursively dividing it at random points and swapping the resulting substrings.

For example, for s1 = "great" and s2 = "rgeat", the string s1 can be transformed into s2 by recursively dividing and swapping the substrings. If such a transformation is possible, return true. If it’s not, return false. The challenge is to use dynamic programming to check all possible transformations.

Examples

Example 1

Input: s1 = "great", s2 = "rgeat"

Output: true

One possible scenario applied on s1 is: "great" --> "gr/eat" // divide at random index. "gr/eat" --> "gr/eat" // random decision is not to swap the two substrings and keep them in order. "gr/eat" --> "g/r / e/at" // apply the same algorithm recursively on both substrings. divide at random index each of them. "g/r / e/at" --> "r/g / e/at" // random decision was to swap the first substring and to keep the second substring in the same order. "r/g / e/at" --> "r/g / e/ a/t" // again apply the algorithm recursively, divide "at" to "a/t". "r/g / e/ a/t" --> "r/g / e/ a/t" // random decision is to keep both substrings in the same order. The algorithm stops now, and the result string is "rgeat" which is s2. As one possible scenario led s1 to be scrambled to s2, we return true.

Example 2

Input: s1 = "abcde", s2 = "caebd"

Output: false

Example details omitted.

Example 3

Input: s1 = "a", s2 = "a"

Output: true

Example details omitted.

Constraints

s1.length == s2.length
1 <= s1.length <= 30
s1 and s2 consist of lowercase English letters.

Solution Approach

State Transition DP

The problem can be solved using dynamic programming by keeping track of valid scrambles at each substring level. We maintain a 3D DP table where dp[i][j][k] indicates if a substring starting at index i of s1 can be scrambled to form a substring starting at index j of s2 with length k.

Memoization to Optimize Computation

Memoization helps store the results of previously computed subproblems to avoid redundant calculations. This is crucial to reducing the time complexity of the solution from O(n^6) to O(n^4), making the algorithm feasible within the problem’s constraints.

Recursive Split and Swap

The string can be recursively divided at various points, and for each division, we either swap or retain the order of the substrings. The recursive nature of the problem is handled by iterating over all possible split points and checking both scenarios: with and without swapping.

Complexity Analysis

Metric	Value
Time	O(n^4)
Space	O(n^3)

The time complexity is O(n^4) due to the nested loops for both substring length and starting positions, along with memoization that helps to avoid redundant recalculations. Space complexity is O(n^3), as we store results for all substrings of s1 and s2 at every level of recursion.

What Interviewers Usually Probe

The candidate demonstrates understanding of dynamic programming by explaining the state transition approach and memoization.
The candidate recognizes the importance of efficiently handling recursive calls using memoization to avoid recomputation.
The candidate may struggle with the recursive nature of the problem if they fail to correctly identify the base cases or transitions.

Common Pitfalls or Variants

Common pitfalls

The candidate may overlook the importance of memoization, leading to a time complexity of O(n^6) instead of O(n^4).
Failing to account for both swap and non-swap scenarios in recursive calls can result in incorrect answers.
Not properly handling edge cases, such as when the strings are already identical or cannot be scrambled into one another, could lead to unnecessary computation.

Follow-up variants

Increasing the length of the strings beyond the current constraints to test the performance of the algorithm.
Modifying the problem to check for scrambled substrings within a larger string rather than the whole string.
Introducing additional characters or larger alphabets in the strings, potentially testing the space and time efficiency of the solution.

FAQ

What is the time complexity of solving the Scramble String problem?

The time complexity is O(n^4) due to the memoization technique and the recursive nature of the problem.

How does dynamic programming help in solving Scramble String?

Dynamic programming helps by storing intermediate results of string scrambles, allowing you to avoid redundant calculations and speeding up the solution process.

What is the main challenge when solving Scramble String?

The main challenge is efficiently managing the recursive calls while checking both swap and non-swap scenarios, which requires careful use of memoization.

Can the Scramble String problem be solved without dynamic programming?

It is possible, but without dynamic programming, the solution would be inefficient, leading to redundant calculations and higher time complexity.

What are some common mistakes when solving Scramble String?

Common mistakes include failing to properly handle recursive cases, neglecting memoization, and missing edge cases like identical or unswappable strings.

terminal

Solution

Solution 1: Memorized Search

We design a function $dfs(i, j, k)$, which means whether the substring starting from $i$ with length $k$ in $s_1$ can be converted into the substring starting from $j$ with length $k$ in $s_2$. If it can be converted, return `true`, otherwise return `false`. The answer is $dfs(0, 0, n)$, where $n$ is the length of the string.

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))

Solution 2: Dynamic Programming (Interval DP)

We define $f[i][j][k]$ as whether the substring of length $k$ starting from $i$ of string $s_1$ can be transformed into the substring of length $k$ starting from $j$ of string $s_2$. Then the answer is $f[0][0][n]$, where $n$ is the length of the string.

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))

Continue Practicing

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