Problem Description

The problem gives us a possible operation on a string where we can take any non-empty string (other than a single character) and split it into two non-empty substrings. We then have the option to either swap these two substrings or leave them as they are. This process can be applied recursively to the new substrings generated from the split. With these rules in mind, we are asked to determine if a given string s2 can be made from another string s1 through one or more of the mentioned scrambling operations, assuming s1 and s2 are of the same length.

Intuition

The core of the solution lies in understanding that the problem can be broken down into smaller subproblems. This is a classic example of a divide and conquer algorithm. Given two strings s1 and s2, our goal is to see if there is a way to split both strings into substrings such that, after potentially swapping the substrings, the smaller pieces are equal or can themselves be further scrambled to be equal.

Taking this intuition, we can design a recursive function that tries to split the strings into two parts in various ways and then checks if after swapping or not swapping, the resulting substrings could be made equal through further recursion. To optimize the recursive solution and eliminate redundant calculations, we use memorization, which stores the results of subproblems that have already been solved.

The function dfs(i, j, k) represents a recursive depth-first search with memorization that checks if the substring of s1 starting at index i and of length k can be transformed into the substring of s2 starting at index j and of the same length. This is done by checking, for each possible split, both cases: keeping the order of the substrings untouched, and swapping them. If any of these cases yields true, we can conclude that s2 is a scrambled version of s1.

To implement this efficiently, we use the cache decorator (from Python's functools module) to memorize the results of subproblems, thus ensuring that our algorithm runs in polynomial time. The recursive function will stop splitting once it reaches substrings of length 1, as those cannot be split further, and a simple character comparison will suffice to continue the recursion. Without memorization, the solution would be prohibitively slow due to the exponential growth of possible splits as the string length increases.

Learn more about Dynamic Programming patterns.

Solution Approach

The solution to this problem utilizes a technique known as memorized recursion, which is a subset of dynamic programming. The idea is that we build our solution from the bottom up by solving all possible subproblems and storing their results to avoid redundant computation for overlapping subproblems.

Here's a breakdown of the implementation:

• Memorized Recursion (dfs function): A depth-first search is performed recursively by the function dfs(i, j, k) which attempts to determine whether a substring of s1 starting at i and having a length k can be transformed into a substring of s2 starting at j of the same length (k). We use Python's @cache decorator to store the results of subproblems in memory, essentially creating a memoization table that the dfs function accesses before computing a new result. This significantly speeds up the algorithm by avoiding recomputation.

• Subproblem Conditions: For each call to dfs(i, j, k), two cases are possible depending on whether the substrings are swapped or not:

  1. When there is no swap, we check if dfs(i, j, h) and dfs(i + h, j + h, k - h) return true for some split length h, meaning that the first h characters of the two substrings and the remaining k - h characters of the two substrings can make a scramble respectively.
  2. When there is a swap, we check if dfs(i, j + k - h, h) and dfs(i + h, j, k - h) return true for some split length h, meaning that the last h characters of the first substring and the first h characters of the second substring, and the remaining characters of both substrings can make a scramble, respectively.

• Base Case: dfs includes a base case for when the length of the substring (k) is 1, at which point it simply compares the individual characters at the given indices i and j in s1 and s2, respectively.

• Return Value: The overall result comes from the initial call to dfs(0, 0, len(s1)), which checks if the entire s1 can be scrambled to result in s2.

• Complexity: The memoization search has a time complexity of O(n^4) and a space complexity of O(n^3), where n is the length of the strings. Time complexity is such because, in the worst case, we may need to check each possible split for each possible substring length k at every possible starting index i and j. Space complexity comes from the amount of information we need to store in our cache.

This recursive and memoization approach ensures we only compute what's necessary, making the problem solvable within a reasonable amount of time for larger strings.

Example Walkthrough

Let's consider a small example to illustrate the solution approach. Suppose we have two strings s1 = "great" and s2 = "rgeat". We want to find out if s2 can be obtained from s1 by applying the scrambling operations described in the problem.

Step 1: Check if s1 can be scrambled to become s2 by calling dfs(0, 0, len(s1)). This is the top-level problem we need to solve.

Step 2: In our dfs(i, j, k) function, we start with i = 0, j = 0, and k = len("great") = 5. We need to check every possible split.

Step 3: Let's split s1 and s2 into h = 2 characters from the start and the remaining k - h = 3 characters. For s1 ("great"), the two parts are "gr" and "eat". For s2 ("rgeat"), the corresponding parts are "rg" and "eat".

In the no-swap case:

• For the first 2 characters, a recursive call dfs(0, 0, 2) checks if "gr" can be scrambled into "rg", which is true. So "gr" can be transformed into "rg" by swapping the characters.
• For the remaining 3 characters, a recursive call dfs(2, 2, 3) checks if "eat" can be scrambled into "eat", which is trivially true.

The recursive calls would return true since both parts can be made to match. Therefore, s2 is a valid scramble of s1 without even needing to check the swap case for this split.

Step 4: The algorithm would then proceed to check all other possible ways of splitting s1 and s2. However, since we have already found a valid scramble, the algorithm doesn't need to continue further for this example.

Step 5: After all the recursion and verification, if any of the recursive calls return true, we conclude that s2 is a scrambled version of s1. Since we found that dfs(0, 0, len(s1)) returns true, s2 = "rgeat" can indeed be made from s1 = "great".

Step 6: During the process, the results of various calls to dfs are stored in the memoization table. Should the same subproblems arise, the algorithm retrieves the result from the table instead of recalculating, effectively reducing computation time and complexity.

In summary, by breaking the problem into smaller subproblems, recursing on those, and memoizing the results, the algorithm can efficiently determine if one string is a scrambled version of the other.

Solution Implementation

Time and Space Complexity

The given Python solution leverages recursion with memoization to determine if one string is a scramble of another. Here's an analysis of its time and space complexities:

Time Complexity:

The dfs function is memoized and explores each possible split of the string segments once for every distinct pair of starting indices (i, j) and length k. Since i and j can each range from 0 to n-1 (for an n-character string) and k can range from 1 to n, there are O(n^2) pairs of starting indices and O(n) possible lengths for each pair. Thus, the time complexity can seem to be O(n^3) considering each possible (i, j, k) triplet is calculated once due to memoization.

However, within every call to dfs(i, j, k), there is a loop that runs k-1 times. Since k can be up to n, the loop can contribute at most O(n) operations per memoized function call. This loop is where the fourth dimension of the time complexity stems from, leading to a total time complexity of O(n^4).

Space Complexity:

The space complexity is primarily dictated by the memoization cache. The space needed for the cache relates directly to the number of distinct arguments passed to the dfs function, which, as explained, is the product of the different values i, j, and k can take. Since i and j range from 0 to n-1 and k ranges from 1 to n, the max number of distinct states stored in the cache is n * n * n, which is O(n^3).

Besides the cache, the space complexity also includes the space for the call stack due to recursion. In the worst case, the recursive calls could stack up to O(n) deep (the maximum possible depth of recursion is when k decreases by 1 each time). However, this does not affect the overall space complexity, which remains O(n^3) due to the dominating size of the memoization cache.

Learn more about how to find time and space complexity quickly using problem constraints.