How do you solve Scramble String on LeetCode?

Scramble String (LeetCode #87) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n³) time complexity and O(n³) space complexity. Key insight: Dynamic programming can significantly reduce redundant calculations.

What is the brute force solution for Scramble String?

The brute force approach for Scramble String is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute-force approach involves generating all possible scrambled versions of the first string and checking if the second string matches any of them. This is straightforward but inefficient for larger strings due to the exponential number of combinations. The optimal Optimal Solution approach improves this to O(n³).

What algorithm or data structure is used to solve Scramble String?

Scramble String uses the following concepts: String, Dynamic Programming. The recommended patterns to study are: Dynamic Programming, Recursion.

What are the interview tips for Scramble String?

Explain your thought process clearly and ask clarifying questions. Consider edge cases, such as strings of length 1 or completely different strings. Practice writing both recursive and iterative solutions to build flexibility in your approach.

What are common mistakes when solving Scramble String?

Not checking for equal lengths early on. Failing to account for character frequency when comparing strings.

#87

Scramble String

Hard

StringDynamic ProgrammingDynamic ProgrammingRecursion

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n³)
Space	O(1)	O(n³)

💡

Intuition

Time O(n³)Space O(n³)

The optimal solution uses dynamic programming to store results of previously computed states, avoiding redundant calculations. This significantly reduces the time complexity by ensuring that each unique combination of substrings is only computed once.

⚙️

Algorithm

4 steps

1Step 1: Create a 3D DP table where dp[i][j][k] indicates whether s1 from index i of length k is a scramble of s2 from index j of length k.
2Step 2: Initialize the DP table for substrings of length 1 (single characters).
3Step 3: Iterate through lengths from 2 to n, and for each length, check all possible splits.
4Step 4: Update the DP table based on whether the substrings can be scrambled by checking both swapping and non-swapping cases.

solution.py16 lines

1def isScramble(s1, s2):
2    n = len(s1)
3    if n != len(s2): return False
4    dp = [[[False] * (n + 1) for _ in range(n)] for _ in range(n)]
5    for i in range(n):
6        for j in range(n):
7            dp[i][j][1] = (s1[i] == s2[j])
8    for length in range(2, n + 1):
9        for i in range(n - length + 1):
10            for j in range(n - length + 1):
11                for split in range(1, length):
12                    if (dp[i][j][split] and dp[i + split][j + split][length - split]) or 
13                       (dp[i][j + length - split][split] and dp[i + split][j][length - split]):
14                        dp[i][j][length] = True
15                        break
16    return dp[0][0][n]

ℹ

Complexity note: The time complexity is O(n³) due to the three nested loops iterating over the length of the string and the possible splits. The space complexity is also O(n³) because of the 3D DP table used to store results.

1Dynamic programming can significantly reduce redundant calculations.
2Understanding the recursive nature of the problem is crucial for both brute force and optimal solutions.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.