Scramble String Leetcode Solution

In this post, we are going to solve the Scramble String Leetcode Solution problem of Leetcode. This Leetcode problem is done in many programming languages like C++, Java, and Python.

Table of Contents

Problem

We can scramble a string s to get a string t using the following algorithm:

If the length of the string is 1, stop.

If the length of the string is > 1, do the following:
- Split the string into two non-empty substrings at a random index, i.e., if the string is s, divide it to x and y where s = x + y.
- Randomly decide to swap the two substrings or to keep them in the same order. i.e., after this step, s may become s = x + y or s = y + x.
- Apply step 1 recursively on each of the two substrings x and y.

Given two strings s1 and s2 of the same length, return true if s2 is a scrambled string of s1, otherwise, return false.

Example 1:

Input: s1 = "great", s2 = "rgeat"
Output: true
Explanation: One possible scenario applied on s1 is:
"great" --> "gr/eat" // divide at random index.
"gr/eat" --> "gr/eat" // random decision is not to swap the two substrings and keep them in order.
"gr/eat" --> "g/r / e/at" // apply the same algorithm recursively on both substrings. divide at random index each of them.
"g/r / e/at" --> "r/g / e/at" // random decision was to swap the first substring and to keep the second substring in the same order.
"r/g / e/at" --> "r/g / e/ a/t" // again apply the algorithm recursively, divide "at" to "a/t".
"r/g / e/ a/t" --> "r/g / e/ a/t" // random decision is to keep both substrings in the same order.
The algorithm stops now, and the result string is "rgeat" which is s2.
As one possible scenario led s1 to be scrambled to s2, we return true.

Example 2:

Input: s1 = "abcde", s2 = "caebd"
Output: false

Example 3:

Input: s1 = "a", s2 = "a"
Output: true

Constraints:

s1.length == s2.length
1 <= s1.length <= 30
s1 and s2 consist of lowercase English letters.

Now, let’s see the leetcode solution of Scramble String Leetcode Solution.

Scramble String Leetcode Solution in Python

class Solution:
  def isScramble(self, s1: str, s2: str) -> bool:
    if s1 == s2:
      return True
    if len(s1) != len(s2):
      return False
    if Counter(s1) != Counter(s2):
      return False

    for i in range(1, len(s1)):
      if self.isScramble(s1[:i], s2[:i]) and self.isScramble(s1[i:], s2[i:]):
        return True
      if self.isScramble(s1[:i], s2[len(s2) - i:]) and self.isScramble(s1[i:], s2[:len(s2) - i]):
        return True

    return False

Scramble String Leetcode Solution in CPP

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

    vector<int> count(128);

    for (int i = 0; i < s1.length(); ++i) {
      ++count[s1[i]];
      --count[s2[i]];
    }

    if (any_of(begin(count), end(count), [](int c) { return c != 0; }))
      return memo[hashedKey] = false;

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

    return memo[hashedKey] = false;
  }

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

Scramble String Leetcode Solution in Java

class Solution {
  public boolean isScramble(String s1, String s2) {
    if (s1.equals(s2))
      return true;
    if (s1.length() != s2.length())
      return false;
    final String hashedKey = s1 + "+" + s2;
    if (memo.containsKey(hashedKey))
      return memo.get(hashedKey);

    int[] count = new int[128];

    for (int i = 0; i < s1.length(); ++i) {
      ++count[s1.charAt(i)];
      --count[s2.charAt(i)];
    }

    for (final int c : count)
      if (c != 0) {
        memo.put(hashedKey, false);
        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))) {
        memo.put(hashedKey, true);
        return true;
      }
      if (isScramble(s1.substring(0, i), s2.substring(s2.length() - i)) &&
          isScramble(s1.substring(i), s2.substring(0, s2.length() - i))) {
        memo.put(hashedKey, true);
        return true;
      }
    }

    memo.put(hashedKey, false);
    return false;
  }

  private Map<String, Boolean> memo = new HashMap<>();
}

Note: This problem Scramble String is generated by Leetcode but the solution is provided by Chase2learn This tutorial is only for Educational and Learning purposes.

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