# Scramble String

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.

## Solution

1. Recursion + pruning.
2. 3-dimensional dp.
• `dp[k][i][j] == true` means string s1(start from i, length k) is a scrambled string of string s2(start from j, length k).

## Code

``````public class Solution {
public boolean isScramble_1(String s1, String s2) {
if (s1.compareTo(s2) == 0) return true;
if (s1.length() != s2.length()) return false;
return isScrambleRe(s1, s2);
}
public boolean isScrambleRe(String s1, String s2) {
if (hasSameLetters(s1, s2) == false) return false;
int len = s1.length();
if (len == 0 || len == 1) return true;
for (int i = 1; i < len; ++i) {
if (isScrambleRe(s1.substring(0,i), s2.substring(0,i))
&& isScrambleRe(s1.substring(i), s2.substring(i))
|| isScrambleRe(s1.substring(0,i), s2.substring(len-i))
&& isScrambleRe(s1.substring(i), s2.substring(0,len-i))) {
return true;
}
}
return false;
}
public boolean hasSameLetters(String s1, String s2) {
if (s1.compareTo(s2) == 0) return true;
if (s1.length() != s2.length()) return false;
int[] count = new int[256];
for (int i = 0; i < s1.length(); ++i) count[(int)s1.charAt(i)]++;
for (int i = 0; i < s2.length(); ++i) count[(int)s2.charAt(i)]--;
for (int i = 0; i < 256; ++i)
if (count[i] != 0) return false;
return true;
}

public boolean isScramble(String s1, String s2) {
if (s1.compareTo(s2) == 0) return true;
if (s1.length() != s2.length()) return false;
int N = s1.length();
boolean[][][] dp = new boolean[N+1][N][N];
for (int k = 1; k <= N; ++k) {
for (int i = 0; i <= N - k; ++i) {
for (int j = 0; j <= N - k; ++j) {
dp[k][i][j] = false;
if (k == 1) dp[1][i][j] = (s1.charAt(i) == s2.charAt(j));
for (int p = 1; p < k && !dp[k][i][j]; ++p) {
if (dp[p][i][j] && dp[k-p][i+p][j+p] || dp[p][i][j+k-p] && dp[k-p][i+p][j])
dp[k][i][j] = true;
}
}
}
}
return dp[N][0][0];
}
}
``````