Given a string s1, we may represent it as a binary tree by partitioning it to two non-empty substrings recursively.

Below is one possible representation of s1 = `"great"`:

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

To scramble the string, we may choose any non-leaf node and swap its two children.

For example, if we choose the node `"gr"` and swap its two children, it produces a scrambled string `"rgeat"`.

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

We say that `"rgeat"` is a scrambled string of `"great"`.

Similarly, if we continue to swap the children of nodes `"eat"` and `"at"`, it produces a scrambled string `"rgtae"`.

```    rgtae
/    \
rg    tae
/ \    /  \
r   g  ta  e
/ \
t   a
```

We say that `"rgtae"` is a scrambled string of `"great"`.

Given two strings s1 and s2 of the same length, determine if s2 is a scrambled string of s1.

Example 1:

```Input: s1 = "great", s2 = "rgeat"
Output: true
```

Example 2:

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

Java:

```public class Solution {
public boolean isScramble(String s1, String s2) {
if (s1.equals(s2)) return true;

int[] letters = new int[26];
for (int i=0; i<s1.length(); i++) {
letters[s1.charAt(i)-'a']++;
letters[s2.charAt(i)-'a']--;
}
for (int i=0; i<26; i++) if (letters[i]!=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;
}
}
```

Python:

```# Time:  O(n^4)
# Space: O(n^3)
class Solution(object):
# @return a boolean
def isScramble(self, s1, s2):
if not s1 or not s2 or len(s1) != len(s2):
return False
if s1 == s2:
return True
result = [[[False for j in xrange(len(s2))] for i in xrange(len(s1))] for n in xrange(len(s1) + 1)]
for i in xrange(len(s1)):
for j in xrange(len(s2)):
if s1[i] == s2[j]:
result[1][i][j] = True

for n in xrange(2, len(s1) + 1):
for i in xrange(len(s1) - n + 1):
for j in xrange(len(s2) - n + 1):
for k in xrange(1, n):
if result[k][i][j] and result[n - k][i + k][j + k] or\
result[k][i][j + n - k] and result[n - k][i + k][j]:
result[n][i][j] = True
break

return result[n][0][0]　　```

C++: Recursion

```class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1==s2)
return true;

int len = s1.length();
int count[26] = {0};
for(int i=0; i<len; i++)
{
count[s1[i]-'a']++;
count[s2[i]-'a']--;
}

for(int i=0; i<26; i++)
{
if(count[i]!=0)
return false;
}

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

C++: Recursion

```class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1.size() != s2.size()) return false;
if (s1 == s2) return true;
string str1 = s1, str2 = s2;
sort(str1.begin(), str1.end());
sort(str2.begin(), str2.end());
if (str1 != str2) return false;
for (int i = 1; i < s1.size(); ++i) {
string s11 = s1.substr(0, i);
string s12 = s1.substr(i);
string s21 = s2.substr(0, i);
string s22 = s2.substr(i);
if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
s21 = s2.substr(s1.size() - i);
s22 = s2.substr(0, s1.size() - i);
if (isScramble(s11, s21) && isScramble(s12, s22)) return true;
}
return false;
}
};　　```

C++: DP

```class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1.size() != s2.size()) return false;
if (s1 == s2) return true;
int n = s1.size();
vector<vector<vector<bool> > > dp (n, vector<vector<bool> >(n, vector<bool>(n + 1, false)));
for (int i = 0; i < n; ++i) {
for (int j = 0; j < n; ++j) {
dp[i][j][1] = s1[i] == s2[j];
}
}
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) {
if ((dp[i][j][k] && dp[i + k][j + k][len - k]) || (dp[i + k][j][len - k] && dp[i][j + len - k][k])) {
dp[i][j][len] = true;
}
}
}
}
}
return dp[0][0][n];
}
};
```

C++:

```class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1.size() != s2.size()) return false;
if (s1 == s2) return true;
int n = s1.size();
vector<vector<vector<bool> > > dp (n, vector<vector<bool> >(n, vector<bool>(n + 1, false)));
for (int i = n - 1; i >= 0; --i) {
for (int j = n - 1; j >= 0; --j) {
for (int k = 1; k <= n - max(i, j); ++k) {
if (s1.substr(i, k) == s2.substr(j, k)) {
dp[i][j][k] = true;
} else {
for (int t = 1; t < k; ++t) {
if ((dp[i][j][t] && dp[i + t][j + t][k - t]) || (dp[i][j + k - t][t] && dp[i + t][j][k - t])) {
dp[i][j][k] = true;
break;
}
}
}
}
}
}
return dp[0][0][n];
}
};
```

# All LeetCode Questions List 题目汇总

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