[Dynamic Programming] Scramble String - Hard

The Scramble String problem is a well-known challenge in dynamic programming. It tests if we can change one string into another by swapping parts of it. Two strings are scramble strings if we can get one from the other with some character swaps. This problem helps us learn dynamic programming better. It also shows us different ways to solve problems like recursion, memoization, and iterative methods.

In this article, we will look at many ways to solve the Scramble String problem using dynamic programming. We will check out recursive solutions, dynamic programming techniques in Python and C++, memoization methods, and iterative approaches. Also, we will compare these different solutions and see how well they perform. We will also answer common questions about the Scramble String problem. The topics we will cover include:

  • Dynamic Programming Approaches for Scramble String Problem
  • Recursive Solution for Scramble String in Java
  • Dynamic Programming Solution for Scramble String in Python
  • Memoization Technique for Scramble String in C++
  • Iterative Approach for Scramble String in Java
  • Optimized Space Complexity Solution for Scramble String in Python
  • Bottom-Up Dynamic Programming for Scramble String in C++
  • Comparative Analysis of Different Solutions for Scramble String
  • Performance Benchmarking of Scramble String Solutions
  • Frequently Asked Questions

By learning these ideas, we can understand how to solve the Scramble String problem better and improve our skills in dynamic programming. If we want to know more about dynamic programming, we can check articles about Fibonacci numbers and the climbing stairs problem. We can look at Dynamic Programming Fibonacci Number and Dynamic Programming Climbing Stairs.

Recursive Solution for Scramble String in Java

We will look at a recursive solution to the Scramble String problem. This problem checks if two strings can change into each other by swapping parts of them. We compare parts of both strings to see if they are equal, letting swaps happen.

Here is the Java code for the recursive solution:

public class ScrambleString {
    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2)) return true;
        if (s1.length() != s2.length()) return false;

        int[] count = new int[26];
        for (char c : s1.toCharArray()) count[c - 'a']++;
        for (char c : s2.toCharArray()) count[c - 'a']--;
        
        for (int i : count) {
            if (i != 0) return false;
        }

        int length = s1.length();
        for (int i = 1; i < length; i++) {
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) && 
                isScramble(s1.substring(i), s2.substring(i))) {
                return true;
            }
            if (isScramble(s1.substring(0, i), s2.substring(length - i)) && 
                isScramble(s1.substring(i), s2.substring(0, length - i))) {
                return true;
            }
        }
        return false;
    }

    public static void main(String[] args) {
        ScrambleString solution = new ScrambleString();
        String s1 = "great";
        String s2 = "rgeat";
        System.out.println(solution.isScramble(s1, s2)); // Output: true
    }
}

Explanation:

We use the method isScramble to see if s1 can become s2.

First, we check if the strings are equal and if they have the same length. Then, we count the characters in both strings to make sure they have the same letters.

Next, we use recursion to check all ways to split the strings. We check both the same order and the swapped order.

This recursive way can take a long time for bigger strings because it has a high time complexity. To make it better, we can use memoization or dynamic programming methods.

For more advanced methods with dynamic programming, you can check the Dynamic Programming Approaches for Scramble String Problem.

Dynamic Programming Solution for Scramble String in Python

We can solve the Scramble String problem well with a dynamic programming method. The main idea is to use a 3D array to keep results of smaller problems. Here, dp[i][j][l] shows if the substring s1[i:i+l] can be scrambled to become s2[j:j+l].

Implementation

Here is the Python code for the dynamic programming solution to the Scramble String problem:

def isScramble(s1: str, s2: str) -> bool:
    n = len(s1)
    if n != len(s2):
        return False
    
    # Create a 3D DP array initialized to False
    dp = [[[False] * (n + 1) for _ in range(n)] for _ in range(n)]
    
    # Base case: single letter substrings
    for i in range(n):
        for j in range(n):
            dp[i][j][1] = (s1[i] == s2[j])
    
    # Fill the DP table
    for length in range(2, n + 1):  # length of the substring
        for i in range(n - length + 1):  # starting index in s1
            for j in range(n - length + 1):  # starting index in s2
                # Check all possible partitions
                for k in range(1, length):  # partition point
                    if (dp[i][j][k] and dp[i + k][j + k][length - k]) or \
                       (dp[i][j + length - k][k] and dp[i + k][j][length - k]):
                        dp[i][j][length] = True
                        break
    
    return dp[0][0][n]

# Example usage
s1 = "great"
s2 = "rgeat"
print(isScramble(s1, s2))  # Output: True

Explanation of the Code

  • Initialization: We create a 3D list dp to track if substrings of s1 can be scrambled to match substrings of s2.
  • Base Case: For substrings of length 1, we simply compare the characters.
  • Filling the DP Table: For each length of substring, we check all ways to split the substrings. Then we update the dp table based on what we found before.
  • Result: The function gives back True if we can scramble the whole string s1 to form s2.

This method using dynamic programming gives us a good way to solve the Scramble String problem. It is faster than using a simple recursive method. If you want to learn more, you can read about dynamic programming techniques to understand these ideas better.

Memoization Technique for Scramble String in C++

The Scramble String problem is about finding if we can change one string into another by swapping non-empty parts of it. We can use a memoization method to make the recursive solution faster. This method saves the results of smaller problems so we do not have to calculate them again.

In C++, we will use a 3D vector to keep the results of our recursive calls. The size of the vector shows the lengths of the parts we are comparing. Let’s see how we can use the memoization technique for the Scramble String problem:

#include <iostream>
#include <vector>
#include <string>
#include <unordered_map>

using namespace std;

class ScrambleString {
public:
    bool isScramble(string s1, string s2) {
        if (s1.length() != s2.length()) return false;
        int n = s1.length();
        vector<vector<vector<int>>> memo(n, vector<vector<int>>(n, vector<int>(n + 1, -1)));
        return dfs(s1, s2, 0, 0, n, memo);
    }

private:
    bool dfs(string &s1, string &s2, int i, int j, int length, vector<vector<vector<int>>> &memo) {
        if (memo[i][j][length] != -1) return memo[i][j][length];

        if (s1.substr(i, length) == s2.substr(j, length)) {
            memo[i][j][length] = true;
            return true;
        }

        for (int k = 1; k < length; k++) {
            if ((dfs(s1, s2, i, j, k, memo) && dfs(s1, s2, i + k, j + k, length - k, memo)) ||
                (dfs(s1, s2, i, j + length - k, k, memo) && dfs(s1, s2, i + k, j, length - k, memo))) {
                memo[i][j][length] = true;
                return true;
            }
        }

        memo[i][j][length] = false;
        return false;
    }
};

int main() {
    ScrambleString solution;
    string s1 = "great";
    string s2 = "rgeat";
    if (solution.isScramble(s1, s2)) {
        cout << s1 << " can be scrambled to " << s2 << endl;
    } else {
        cout << s1 << " cannot be scrambled to " << s2 << endl;
    }
    return 0;
}

Explanation of the Code:

  • Data Structures: We use a 3D vector called memo. It saves the results. memo[i][j][length] tells us if the part of s1 starting at i with length length can be scrambled to match the part of s2 starting at j.
  • Base Case: If the parts are the same, we save true in memo and return.
  • Recursive Cases: We check possible split points. We look at both ways of arranging the parts.
  • Efficiency: Using memoization helps a lot. It cuts down the number of recursive calls. This makes our time complexity better from exponential to polynomial.

This method helps us solve the Scramble String problem in C++ efficiently. To learn more about related dynamic programming methods, we can check the article on Dynamic Programming Edit Distance.

Iterative Approach for Scramble String in Java

We can use the iterative way to solve the Scramble String problem. This method uses dynamic programming to check if one string is a scrambled version of another. We avoid the extra work of recursion. Instead, we use a 3D boolean array to keep track of our results.

Code Implementation

Here is the Java code for the iterative solution:

public class ScrambleString {
    public boolean isScramble(String s1, String s2) {
        int n = s1.length();
        if (n != s2.length()) return false;
        
        // DP table
        boolean[][][] dp = new boolean[n][n][n + 1];

        // Initialize for strings of length 1
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j][1] = s1.charAt(i) == s2.charAt(j);
            }
        }

        // Check substrings of length 2 to n
        for (int len = 2; len <= n; len++) {
            for (int i = 0; i <= n - len; i++) {
                for (int j = 0; j <= n - len; j++) {
                    for (int k = 1; k < len; k++) {
                        // Check if s1[i:i+len] is a scramble of s2[j:j+len]
                        if ((dp[i][j][k] && dp[i + k][j + k][len - k]) || 
                            (dp[i][j + len - k][k] && dp[i + k][j][len - k])) {
                            dp[i][j][len] = true;
                            break;
                        }
                    }
                }
            }
        }
        return dp[0][0][n];
    }

    public static void main(String[] args) {
        ScrambleString scrambleString = new ScrambleString();
        String s1 = "great";
        String s2 = "rgeat";
        System.out.println(scrambleString.isScramble(s1, s2)); // Output: true
    }
}

Explanation of the Code

  • 3D DP Array: The dp[i][j][len] means if the substring s1[i:i+len] is a scramble of s2[j:j+len].
  • Initialization: We set the base case for substrings of length 1 in the DP table.
  • Nested Loops: The outer loops go through lengths and starting points. The innermost loop checks all ways to divide the substrings.
  • Result: The final answer is in dp[0][0][n]. This shows if the whole string s1 is a scramble of s2.

This iterative way is much better than the recursive way. It works faster for bigger strings. For more about dynamic programming, we can read other articles like Dynamic Programming Edit Distance.

Optimized Space Complexity Solution for Scramble String in Python

We can solve the Scramble String problem in Python using a method called dynamic programming. This helps us use less space by only needing two layers of a 3D list instead of a full 3D array. The goal is to find out if one string can change into another by swapping parts of it.

Algorithm

  1. Definition: We define dp[i][j][k] as a true or false value. It shows if the part of s1 from i to i+k can change into the part of s2 from j to j+k.
  2. Initialization: For parts of length 1, we compare the characters directly.
  3. Recursion: For longer parts, we check two cases:
    • Without swapping: dp[i][j][k] = dp[i][j][m] and dp[i+m][j+m][k-m]
    • With swapping: dp[i][j][k] = dp[i][j+k-m][m] and dp[i+m][j][k-m]

Implementation

Here is the Python code for this solution:

def isScramble(s1: str, s2: str) -> bool:
    if len(s1) != len(s2):
        return False
    n = len(s1)

    # Create a 3D array with only two layers to save space
    dp = [[[False] * (n + 1) for _ in range(n)] for _ in range(2)]

    # Fill the base case for length 1 substrings
    for i in range(n):
        for j in range(n):
            dp[0][i][j] = (s1[i] == s2[j])

    # Iterate over substring lengths
    for length in range(2, n + 1):  # length of substring
        for i in range(n - length + 1):
            for j in range(n - length + 1):
                for m in range(1, length):  # split point
                    if (dp[(length - 2) % 2][i][j] and dp[(length - 2) % 2][i + m][j + m]) or \
                       (dp[(length - 2) % 2][i][j + m] and dp[(length - 2) % 2][i + m][j]):
                        dp[length % 2][i][j] = True
                        break
                # Reset the previous layer
                if length > 2:
                    dp[(length - 2) % 2][i][j] = False

    return dp[(n) % 2][0][0]

# Example usage
s1 = "great"
s2 = "rgeat"
print(isScramble(s1, s2))  # Output: True

Explanation

We reduce the space needed to O(n^2) by using only two layers of the 3D DP array. This is enough for our calculations. The function checks if the two strings can change to match each other. It does this by checking both cases of swapping and not swapping parts. This method works well without using too much memory, so it is good for bigger strings.

For more insights into dynamic programming strategies, we can check the Dynamic Programming - Edit Distance article, it may help.

Bottom-Up Dynamic Programming for Scramble String in C++

We can solve the Scramble String problem well by using a bottom-up dynamic programming method. This way, we build a table. This table will show if two substrings are scramble strings of each other.

Algorithm

  1. Initialization: We create a 3D boolean array dp. Here, dp[i][j][k] tells us if the substring s1[i...i+k-1] is a scramble string of s2[j...j+k-1].

  2. Base Case: For substrings with length 1, if the characters are the same, we set dp[i][j][1] = true.

  3. Filling the DP Table:

    • We go through lengths from 2 to n.
    • For each starting index i for s1 and j for s2, we check all split points.
    • We update the dp table based on the scramble string rules.

  4. Return the result: The answer is in dp[0][0][n]. This shows if s1 can be scrambled to make s2.

C++ Implementation

Here is a sample implementation in C++:

#include <vector>
#include <string>

class Solution {
public:
    bool isScramble(std::string s1, std::string s2) {
        int n = s1.size();
        if (n != s2.size()) return false;

        // DP table
        std::vector<std::vector<std::vector<bool>>> dp(n, 
            std::vector<std::vector<bool>>(n, std::vector<bool>(n + 1, false)));

        // Base case
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                dp[i][j][1] = (s1[i] == s2[j]);
            }
        }

        // Fill the DP table
        for (int len = 2; len <= n; len++) {
            for (int i = 0; i <= n - len; i++) {
                for (int j = 0; j <= n - len; j++) {
                    for (int split = 1; split < len; split++) {
                        if ((dp[i][j][split] && dp[i + split][j + split][len - split]) || 
                            (dp[i][j + len - split][split] && dp[i + split][j][len - split])) {
                            dp[i][j][len] = true;
                        }
                    }
                }
            }
        }

        return dp[0][0][n];
    }
};

Key Properties

  • Time Complexity: The time complexity is (O(n^4)). This is because of the nested loops over length, starting points, and split points.
  • Space Complexity: The space complexity is (O(n^3)). This is for the 3D DP array.

This bottom-up dynamic programming solution helps us find out if one string is a scrambled version of another. We build the solution step by step using results we have already found.

Comparative Analysis of Different Solutions for Scramble String

We can solve the Scramble String problem using different methods. Each method has its own pros and cons. We will look at recursive, dynamic programming, memoization, and iterative approaches. We will compare them based on time complexity, space complexity, and how easy they are to use.

1. Recursive Solution

  • Time Complexity: Exponential (O(2^n))
  • Space Complexity: O(n) (because of recursion stack)
  • Description: This simple method checks if two strings are scramble strings. It does this by dividing the strings and checking the rules again.
public boolean isScramble(String s1, String s2) {
    if (s1.equals(s2)) return true;
    if (s1.length() != s2.length()) return false;

    int[] count = new int[26];
    for (char c : s1.toCharArray()) count[c - 'a']++;
    for (char c : s2.toCharArray()) {
        if (--count[c - 'a'] < 0) return false;
    }

    for (int i = 1; i < s1.length(); i++) {
        if (isScramble(s1.substring(0, i), s2.substring(0, i)) &&
            isScramble(s1.substring(i), s2.substring(i))) {
            return true;
        }
        if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i)) &&
            isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) {
            return true;
        }
    }
    return false;
}

2. Dynamic Programming Solution

  • Time Complexity: O(n^4)
  • Space Complexity: O(n^3)
  • Description: This method uses a 3D DP array. It saves results of smaller problems. This way, it does not calculate the same thing over and over.
def isScramble(s1: str, s2: str) -> bool:
    n = len(s1)
    if n != len(s2): return False
    dp = [[[False] * n for _ in range(n)] for _ in range(n + 1)]

    for i in range(n):
        for j in range(n):
            dp[1][i][j] = s1[i] == s2[j]

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            for j in range(n - length + 1):
                for k in range(1, length):
                    if (dp[k][i][j] and dp[length - k][i + k][j + k]) or \
                       (dp[k][i][j + length - k] and dp[length - k][i + k][j]):
                        dp[length][i][j] = True
                        break
    return dp[n][0][0]

3. Memoization Technique

  • Time Complexity: O(n^4)
  • Space Complexity: O(n^3)
  • Description: This method improves the recursive solution. It saves results of smaller problems to avoid doing the same work again.
#include <unordered_map>
#include <string>
using namespace std;

class Solution {
public:
    bool isScramble(string s1, string s2) {
        if (s1 == s2) return true;
        if (s1.length() != s2.length()) return false;
        string key = s1 + "," + s2;
        if (memo.count(key)) return memo[key];

        int count[26] = {0};
        for (char c : s1) count[c - 'a']++;
        for (char c : s2) if (--count[c - 'a'] < 0) return memo[key] = false;

        int n = s1.length();
        for (int i = 1; i < n; i++) {
            if ((isScramble(s1.substr(0, i), s2.substr(0, i)) && 
                 isScramble(s1.substr(i), s2.substr(i))) || 
                (isScramble(s1.substr(0, i), s2.substr(n - i)) && 
                 isScramble(s1.substr(i), s2.substr(0, n - i)))) {
                return memo[key] = true;
            }
        }
        return memo[key] = false;
    }

private:
    unordered_map<string, bool> memo;
};

4. Iterative Approach

  • Time Complexity: O(n^4)
  • Space Complexity: O(n^3)
  • Description: This method uses a bottom-up way. It builds results step by step using a DP table.
def isScramble(s1: str, s2: str) -> bool:
    n = len(s1)
    if n != len(s2): return False
    dp = [[[False] * n for _ in range(n)] for _ in range(n + 1)]

    for i in range(n):
        for j in range(n):
            dp[1][i][j] = (s1[i] == s2[j])

    for length in range(2, n + 1):
        for i in range(n - length + 1):
            for j in range(n - length + 1):
                for k in range(1, length):
                    if (dp[k][i][j] and dp[length - k][i + k][j + k]) or \
                       (dp[k][i][j + length - k] and dp[length - k][i + k][j]):
                        dp[length][i][j] = True
                        break
    return dp[n][0][0]

Summary of Analysis

  • Recursive: Easy to write but slow for big strings because of high time complexity.
  • Dynamic Programming: Fast with polynomial time. It needs a lot of space.
  • Memoization: Easy like recursion but faster because it saves results.
  • Iterative: Does not use recursion limits but can be hard to write.

For more reading about dynamic programming, you can check articles on Dynamic Programming Fibonacci Number and Dynamic Programming Edit Distance.

Performance Benchmarking of Scramble String Solutions

When we look at how different solutions work for the Scramble String problem, it is important to think about time and space. The main ways to solve this problem are recursive, dynamic programming (DP), memoization, and iterative methods. Here is a simple comparison of their performance.

Recursive Solution

  • Time Complexity: Exponential, O(2(n2)), where n is the length of the string.
  • Space Complexity: O(n), because of the recursion stack.

Memoization Technique

  • Time Complexity: O(n^3), since we check all possible pairs of substrings.
  • Space Complexity: O(n^2), for saving results in a 3D array.

Java Implementation:

public class ScrambleString {
    private HashMap<String, Boolean> memo = new HashMap<>();

    public boolean isScramble(String s1, String s2) {
        if (s1.equals(s2)) return true;
        String key = s1 + "," + s2;
        if (memo.containsKey(key)) return memo.get(key);

        int[] count = new int[26];
        for (int i = 0; i < s1.length(); i++) {
            count[s1.charAt(i) - 'a']++;
            count[s2.charAt(i) - 'a']--;
        }
        for (int c : count) if (c != 0) return memo.put(key, false);

        for (int i = 1; i < s1.length(); i++) {
            if (isScramble(s1.substring(0, i), s2.substring(0, i)) &&
                isScramble(s1.substring(i), s2.substring(i))) {
                return memo.put(key, true);
            }
            if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i)) &&
                isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) {
                return memo.put(key, true);
            }
        }
        return memo.put(key, false);
    }
}

Dynamic Programming Solution

  • Time Complexity: O(n^4), where n is the length of the strings.
  • Space Complexity: O(n^3), for the 3D DP array.

Python Implementation:

def isScramble(s1: str, s2: str) -> bool:
    n = len(s1)
    dp = [[[False] * n for _ in range(n)] for _ in range(n + 1)]
    
    for i in range(n):
        for j in range(n):
            dp[1][i][j] = s1[i] == s2[j]
    
    for length in range(2, n + 1):
        for i in range(n - length + 1):
            for j in range(n - length + 1):
                for k in range(1, length):
                    if (dp[k][i][j] and dp[length - k][i + k][j + k]) or \
                       (dp[k][i][j + length - k] and dp[length - k][i + k][j]):
                        dp[length][i][j] = True
                        break
    return dp[n][0][0]

Iterative Approach

  • Time Complexity: O(n^4).
  • Space Complexity: O(n^3).

This method builds the solution step by step using a 3D array like the dynamic programming solution.

Performance Benchmarking

To test these solutions, we can run each one on different string lengths and check how long they take to run. The recursive method is not very efficient because it grows exponentially in time. On the other hand, memoization and dynamic programming methods are faster but use more space.

For more information on dynamic programming and related problems, we can check out Dynamic Programming Fibonacci Number or Dynamic Programming Edit Distance.

Frequently Asked Questions

1. What is the Scramble String problem in dynamic programming?

The Scramble String problem is about figuring out if we can change one string into another by swapping parts of it. We do this by looking at smaller parts of the string and checking if we can mix them up. This problem helps us learn about recursion and how to think about combinations. We can make our solutions faster by using memoization or tabulation. This makes it better than just using a simple recursive way.

2. How can I implement the Scramble String solution in Java?

To solve the Scramble String problem in Java, we can use recursion along with memoization. This means we save the results we already calculated in a hashmap or a 2D array. This way, we do not repeat the same work. The recursive function will check both direct changes and scrambled changes of parts of the strings. This helps us run the solution more efficiently.

3. What dynamic programming techniques are used for solving the Scramble String problem?

For the Scramble String problem, we often use memoization and bottom-up tabulation. Memoization saves results that we have found so we do not have to calculate them again. The bottom-up way builds our solution step by step from smaller problems. This helps us save time and space when we solve the problem.

4. How does the iterative approach for Scramble String work in Python?

In Python, the iterative way to solve the Scramble String problem uses a 3D list. This list helps us keep track of if the scrambled states are valid. We fill this list carefully by comparing parts of the strings and using results we calculated before. This way, we can see if one string can become another without using recursion.

5. Can I optimize the space complexity of the Scramble String solution in C++?

Yes, we can make the space usage better in the Scramble String solution in C++. Instead of using a 3D array, we can just use a 2D array. This way, we only keep track of the current and the previous states. It helps us use less memory but still keeps our solution efficient. We need to be careful with the indices when we do the calculations.

For more insights into dynamic programming techniques, check out our articles on Dynamic Programming: Fibonacci Number and Dynamic Programming: Edit Distance.