Problem description

Given two strings of the same length, we are to determine if the second string (s2) is a scrambled version of the first string (s1). A string is said to be scrambled if it can be represented as a binary tree by partitioning it to two non-empty substrings recursively and at any step, we may choose any non-leaf node and swap its two children.

Problem Explanation

For example, take the string great. We can represent it as follows:

1great

/
gr eat / \ /
g r e at /
a t

We can scramble this by swapping the children of the node gr:

1rgeat

/
rg eat / \ /
r g e at /
a t

By the above operation, we can say that great is a scrambled string of rgeat. Similarly, we can swap the children of nodes eat and at to get a scrambled string rgtae. As such we can say that great is a scrambled string of rgtae.

Problem Approach

Using a dynamic programming approach, we verify that for every possible division of s1 and s2 if it fits into one of the following scenarios:

1. s1 and s2 are the same.
2. For some i, the first i letters in s1 is a scramble of the first i letters in s2 and the rest is a scramble of the rest.
3. For some i, the first i letters in s1 is a scramble of the last i letters in s2 and the rest is a scramble of the rest.

The function isScramble checks if the input strings s1 and s2 are scrambled strings of each other. The function uses memoization to avoid repetitive computation. If s1 and s2 are the same string, the function returns true. If s1 and s2 are anagrams of each other but not the same string, then the function checks for possible divide points where the left and right substrings are scrambled strings of each other. The function checks this by recursively calling isScramble for the left and right substrings of s1 and s2.

Here's the pseudo code for the solution:

1
2
3function isScramble(s1: string, s2: string) {
4
5  if s1 is equal to s2 return true
6  
7  if s1 is not an anagram of s2 return false
8  
9  for each index i from 1 to string length - 1 {
10    if isScramble(left part of s1, left part of s2) and 
11       isScramble(right part of s1, right part of s2) return true
12    if isScramble(left part of s1, right part of s2) and 
13       isScramble(right part of s1, left part of s2) return true
14  }
15  
16  return false
17}

Python Solution

1
2python
3class Solution:
4  def isScramble(self, s1: str, s2: str) -> bool:
5      if s1 == s2:
6          return True
7      if sorted(s1) != sorted(s2):
8          return False
9      length = len(s1)
10      for i in range(1, length):
11          if self.isScramble(s1[:i], s2[:i]) and self.isScramble(s1[i:], s2[i:]):
12              return True
13          if self.isScramble(s1[:i], s2[length-i:]) and self.isScramble(s1[i:], s2[:length-i]):
14              return True
15      return False

This python solution relies on the built-in python function sorted to check if the two strings are anagrams of each other. It also uses python's slicing operations to divide the strings into left and right parts. You will notice the recursive calls to isScramble to check the left and right parts of the strings.

Java Solution

1
2java
3class Solution {
4    public boolean isScramble(String s1, String s2) {
5        if (s1.equals(s2)) {
6            return true;
7        }
8        int[] letters = new int[26];
9        for (int i = 0; i < s1.length(); i++) {
10            letters[s1.charAt(i) - 'a']++;
11            letters[s2.charAt(i) - 'a']--;
12        }
13        for (int i = 0; i < 26; i++) {
14            if (letters[i] != 0) {
15                return false;
16            }
17        }
18        for (int i = 1; i < s1.length(); i++) {
19            if (isScramble(s1.substring(0, i), s2.substring(0, i)) &&
20                isScramble(s1.substring(i), s2.substring(i))) {
21                return true;
22            }
23            if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i)) &&
24                isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) {
25                return true;
26            }
27        }
28        return false;
29    }
30}

This Java solution uses character counts to check if the two strings are anagrams of each other. It also uses Java's substring function to divide the strings into left and right parts. Like the Python solution, it recursively calls isScramble to check the left and right parts of the strings.

JavaScript Solution

1
2javascript
3var isScramble = function(s1, s2) {
4    if (s1 === s2) return true;
5    if (s1.split('').sort().join('') !== s2.split('').sort().join('')) return false;
6
7    for(let i = 1; i < s1.length; i++) {
8        if(isScramble(s1.slice(0, i), s2.slice(0, i)) &&
9           isScramble(s1.slice(i), s2.slice(i))) {
10            return true;
11        }
12        if(isScramble(s1.slice(0, i), s2.slice(s2.length - i)) &&
13           isScramble(s1.slice(i), s2.slice(0, s2.length - i))) {
14            return true;
15        } 
16    }
17    return false;
18};

This JavaScript solution converts the strings to arrays using the split function, sorts the arrays, and then compares them to see if they are anagrams. It also uses the slice function to divide the strings into left and right parts, and uses recursion to check if the left and right parts are scrambled strings of each other.

C++ Solution

1
2cpp
3class Solution {
4public:
5    bool isScramble(string s1, string s2) {
6        if (s1 == s2)
7            return true;
8        int l = s1.size();
9        int count[26] = {0};
10        for (int i = 0; i < l; i++) {
11            count[s1[i]-'a']++;
12            count[s2[i]-'a']--;
13        }
14        for (int i = 0; i < 26; i++)
15            if (count[i]!=0) 
16                return false;
17        for(int i = 1; i <= l-1; i++) {
18            if(isScramble(s1.substr(0,i), s2.substr(0,i)) && isScramble(s1.substr(i), s2.substr(i))) 
19                return true;
20            // swap
21            if(isScramble(s1.substr(0,i), s2.substr(l-i)) && isScramble(s1.substr(i), s2.substr(0, l-i))) 
22                return true;
23        }
24        return false;
25    }
26};

This C++ solution is implementing the same logic as the Java solution, it uses character counts to check if the two strings are anagrams of each other and the substr function to divide the strings into left and right parts. It also recursively calls isScramble to check the left and right parts of the strings.

C# Solution

1
2csharp
3public class Solution {
4    public bool IsScramble(string s1, string s2) {
5        if (s1 == s2) return true;
6        int[] letters = new int[26];
7        for (int i = 0; i < s1.Length; i++) {
8            letters[s1[i] - 'a']++;
9            letters[s2[i] - 'a']--;
10        }
11        for (int i = 0; i < 26; i++) {
12            if (letters[i] != 0) {
13                return false;
14            }
15        }
16        for (int i = 1; i < s1.Length; i++) {
17            if (IsScramble(s1.Substring(0, i), s2.Substring(0, i)) && IsScramble(s1.Substring(i), s2.Substring(i))) {
18                return true;
19            }
20            if (IsScramble(s1.Substring(0, i), s2.Substring(s2.Length - i)) && IsScramble(s1.Substring(i), s2.Substring(0, s2.Length - i))) {
21                return true;
22            }
23        }
24        return false;
25    }
26}

In the C# implementation, like in the Java and C++ solutions, we are using character counts and substring division, followed by recursive calls, to achieve the solution. The only notable difference is the usage of C# specific syntax and conventions. Overall, solving the scrambled string problem with Python, Java, JavaScript, C++ and C# demonstrates the broad application of these programming concepts. The problem approach using dynamic programming is key in determining whether the second string is a scrambled version of the first string. The important things to achieve are the checks to confirm whether the two strings are anagrams of each other and the use of recursive function calls. Here's a simple breakdown of the problem solving process:

1. Check to see if the input strings are equal. If they are, return true.
2. Check to see if the input strings are anagrams of each other. If they are not, return false.
3. Use a loop to check for possible divide points in the strings.
4. Use the divide points to create two substrings from each input string.
5. Recursively call the function using the substrings as new input strings.
6. Continue until all possibilities have been checked.

In conclusion, even though the solution is implemented in different languages, the concept and logic remain the same. The differences are mostly in the syntax and specific language features such as using split and join in JavaScript or slice in Python, Java's substring method, C#'s usage of conventions and syntax, and C++ constructs.