Scramble String

To solve this coding challenge of determining whether one string is a scrambled version of another, we can use a dynamic programming-based approach. The main idea revolves around understanding if the given string

s1

can be transformed into

s2

through recursive any-and-all mixes after splitting it at all possible points.

# Explanation

The problem is about determining if one string (

s2

) can be formed by scrambling another string (

s1

). A scrambled string is obtained by:

1. Splitting the string into two non-empty substrings.
2. Either swapping or not swapping these substrings.
3. Recursively applying this process to both substrings.
The constraints help to simplify some aspects:
4. Both strings are of the same length.
5. Both strings consist only of lowercase English letters.
The solution involves:
6. Base cases : Direct comparison when the strings are identical or when their sorted versions do not match.
7. Recursive Check : For possible splits, verifying if either swapping or not swapping can produce the desired string
   s2
   .
8. Dynamic Programming : To avoid redundant calculations, we use a table (
   dp
   ) to store whether substrings of given lengths from both
   s1
   and
   s2
   are scramble equivalents.

Detailed Steps in Pseudocode

Initialization

9. Define a function
   is_scramble
   that uses dynamic programming to solve the problem.
10. Create a 3D DP table
    dp
    where
    dp[length-1][i][j]
    indicates whether the substrings
    s1[i:i+length]
    and
    s2[j:j+length]
    are scramble-equivalent.

Base Cases

11. If the strings are identical, return
    True
    .
12. If the strings are of different lengths or their sorted versions don't match, return
    False
    .

Dynamic Programming

13. Initialize the
    dp
    table for substrings of length 1.
14. For each possible substring length, iterate through all possible starting indices for
    s1
    and
    s2
    .
15. For each possible split point within the substring, update the
    dp
    table based on subproblem results (considering both swapping and not swapping the substrings).

Final Check

1. Return the value from the DP table that corresponds to the entire strings of
   s1
   and
   s2
   .

Here's the pseudocode:

                                            
Function isScramble(s1, s2):

    # Base cases: if s1 is equal to s2, we can return true immediately
    if s1 == s2:
        return True

    # If the strings are not of the same length or their sorted versions do not match, return False
    if len(s1) != len(s2) or sorted(s1) != sorted(s2):
        return False

    # Initialize a 3D DP table (dp) to track scrambling - dp[length][i][j]
    n = len(s1)
    dp = [[[False] * n for _ in range(n)] for _ in range(n)]

    # Fill the base case: single character comparison
    for i in range(n):
        for j in range(n):
            dp[0][i][j] = (s1[i] == s2[j])

    # Check all possible substrings lengths from 2 to n
    for length in range(2, n + 1):
        for i in range(n - length + 1): # Starting index i in s1
            for j in range(n - length + 1): # Starting index j in s2
                # Check all possible split points
                for k in range(1, length):
                    if (dp[k - 1][i][j] and dp[length - k - 1][i + k][j + k]) or \
                       (dp[k - 1][i][j + length - k] and dp[length - k - 1][i + k][j]):
                        dp[length - 1][i][j] = True
                        break

    # The result for the entire strings is stored at dp[n-1][0][0]
    return dp[n - 1][0][0]

# Main function to call the scrambling check
Function main(s1, s2):
    # Invoke the scrambling check
    return isScramble(s1, s2)

                                        

Explanation of Pseudocode

• The function
  isScramble
  manages all steps to determine if
  s1
  can be scrambled to form
  s2
  .
• The conditionals first handle easy cases where we can immediately return True or False.
• The 3D DP table
  dp
  stores results for substrings being equivalent.
• The nested loops fill in this table by looking at all possible substrings lengths and their possible splitting points.
• Finally, the result for
  dp[n-1][0][0]
  gives the answer whether
  s1
  and
  s2
  are scramble equivalents.