# Question

Formatted question description: https://leetcode.ca/all/87.html

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

1. If the length of the string is 1, stop.
2. 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.

# Algorithm

## DFS

To determine if two strings s1 and s2 are scramble, there must exist a length l1 in s1 such that it can be divided into s11 and s12. Similarly, there must exist s21 and s22 such that either s11 and s21 are scramble and s12 and s22 are scramble, or s11 and s22 are scramble and s12 and s21 are scramble.

Consider the example of “rgeat” and “great”. “rgeat” can be divided into “rg” and “eat”, and “great” can be divided into “gr” and “eat”. Since “rg” and “gr” are scrambled and “eat” and “eat” are identical, “rgeat” and “great” are scrambled strings.

## DP

To use dynamic programming for this problem, create a three-dimensional array dp[i][j][n], where i represents the starting character of s1, j represents the starting character of s2, and n represents the current string length. dp[i][j][len] indicates whether the substring of s1 and s2 with length len starting from i and j, respectively, are scrambles of each other.

To determine dp[i][j][len], consider the historical information available. Any string with length less than n has already been determined. Divide the current s1[i...i+len-1] string into two parts, and then evaluate the following two cases:

• The first case is where the left part of s1[i...i+len-1] is a scramble of s2[j...j+len-1] and the right part of s1[i...i+len-1] is a scramble of s2[j...j+len-1].
• The second case is where the left part of s1[i...i+len-1] is a scramble of s2[j+len-k...j+len-1] and the right part of s1[i...i+len-1] is a scramble of s2[j...j+len-k-1].

If either of the two cases is true, then s1[i...i+len-1] and s2[j...j+len-1] are scramble.

There are len-1 possible splitting methods for s1[i...i+len-1]. If any of these splitting methods yield a true value for the state transition equation, then the two strings are scramble. The state transition equation is:

dp[i][j][len] = || (dp[i][j][k] && dp[i+k][j+k][len-k] || dp[i][j+len-k][k] && dp[i+k][j][len-k])

# Code

• 
public class Scramble_String {

public class Solution {
public boolean isScramble(String s1, String s2) {
if (s1 == null || s2 == null || s1.length() != s2.length()) {
return false;
}

int len = s1.length();

// dp[i][j], i,j not index, substring i to j of s1/s2 are scrambled
boolean[][][] dp = new boolean[len][len][len + 1];

for (int i = 0; i < len; i++) {
for (int j = 0; j < len; j++) {
dp[i][j][1] = s1.charAt(i) == s2.charAt(j);
}
}

for (int n = 1; n < len + 1; n++) {
for (int i = 0; i + n - 1 < len; i++) { // @note: here boundary check, I missed
for (int j = 0; j + n - 1 < len; j++) {

for (int sublen = 1; sublen < n; sublen++) {

// same strings: s1="abcde", s2="abcde"
boolean ifSame = dp[i][j][sublen] && dp[i + sublen][j + sublen][n - sublen];

// swapped strings: s1="abcde", s2="cdeab"
// @note: especially for the index here, using an example for assistance
boolean ifSwapped = dp[i][j + n - sublen][sublen] && dp[i + sublen][j][n - sublen];

// dp[i][j][n] = ifSame || ifSwapped; // @note: stupid...being overwritten
if (ifSame || ifSwapped) {
dp[i][j][n] = true;
break;
}
}
}
}
}

return dp[0][0][len];

}
}

public class Solution_over_time { // clearly a lot of repeated calculation during recursion
public boolean isScramble(String s1, String s2) {
if (s1 == null || s2 == null || s1.length() != s2.length()) {
return false;
}

if (s1.length() == 0) {
return s2.length() == 0;
}

if (s1.length() == 1) {
return s1.charAt(0) == s2.charAt(0);
}

// basically swap happens for a node of 2 children
if (s1.length() == 2) {
boolean isSame = s1.equals(s2);
boolean isSwapped = s1.charAt(0) == s2.charAt(1) && s1.charAt(1) == s2.charAt(0);

return isSame || isSwapped;
}

for (int i = 1; i < s1.length(); i++) {

// @note: left and right child chould NOT be empty
// int splitIndex = i;
boolean isThisScramble = isScramble(s1.substring(0, i), s2.substring(0, i))
&& isScramble(s1.substring(i), s2.substring(i));
if (isThisScramble) {
return true;
}
}

return false;

}
}

}

############

class Solution {
public boolean isScramble(String s1, String s2) {
int n = s1.length();
boolean[][][] dp = new boolean[n][n][n + 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);
}
}
for (int len = 2; len <= n; ++len) {
for (int i1 = 0; i1 <= n - len; ++i1) {
for (int i2 = 0; i2 <= n - len; ++i2) {
for (int i = 1; i < len; ++i) {
if (dp[i1][i2][i] && dp[i1 + i][i2 + i][len - i]) {
dp[i1][i2][len] = true;
break;
}
if (dp[i1][i2 + len - i][i] && dp[i1 + i][i2][len - i]) {
dp[i1][i2][len] = true;
break;
}
}
}
}
}
return dp[0][0][n];
}
}

• // OJ: https://leetcode.com/problems/scramble-string
// Time: O(N^(N+1))
// Space: O(N^(N+1))
class Solution {
public:
bool isScramble(string s1, string s2) {
if (s1.size() != s2.size()) return false;
if (s1 == s2) return true;
for (int i = 1; i < s1.size(); ++i) {
string l1 = s1.substr(0, i), r1 = s1.substr(i);
string l2 = s2.substr(0, i), r2 = s2.substr(i);
string r2r = s2.substr(0, s1.size() - i), l2r = s2.substr(s1.size() - i);
if ((isScramble(l1, l2) && isScramble(r1, r2))
|| (isScramble(l1, l2r) && isScramble(r1, r2r))) return true;
}
return false;
}
};

• class Solution:
def isScramble(self, s1: str, s2: str) -> bool:
n = len(s1)
dp = [[[False] * (n + 1) for _ in range(n)] for _ in range(n)]
for i in range(n):
for j in range(n):
dp[i][j][1] = s1[i] == s2[j]
for l in range(2, n + 1):
for i1 in range(n - l + 1):
for i2 in range(n - l + 1):
for i in range(1, l):
if dp[i1][i2][i] and dp[i1 + i][i2 + i][l - i]:
dp[i1][i2][l] = True
break
if dp[i1][i2 + l - i][i] and dp[i1 + i][i2][l - i]:
dp[i1][i2][l] = True
break
return dp[0][0][n]

############

class Solution(object):
def isScramble(self, s1, s2):
"""
:type s1: str
:type s2: str
:rtype: bool
"""
n = len(s1)
m = len(s2)
if sorted(s1) != sorted(s2):
return False

if n < 4 or s1 == s2:
return True

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


• public class Solution {
public bool IsScramble(string s1, string s2) {
if (s1.Length != s2.Length) return false;
var length = s1.Length;
if (length == 0) return true;
var f = new bool[length + 1, length, length];
for (var i = 0; i < length; ++i)
{
for (var j = 0; j < length; ++j)
{
f[1, i, j] = s1[i] == s2[j];
}
}
for (var i = 2; i <= length; ++i)
{
for (var j = 0; j <= length - i; ++j)
{
for (var k = 0; k <= length - i; ++k)
{
for (var l = 1; l < i; ++l)
{
if (f[l, j, k] && f[i - l, j + l, k + l]
|| f[l, j, k + i - l] && f[i - l, j + l, k])
{
f[i, j, k] = true;
break;
}
}
}
}
}

return f[length, 0, 0];
}
}

• func isScramble(s1 string, s2 string) bool {
n := len(s1)
dp := make([][][]bool, n+1)
for i := range dp {
dp[i] = make([][]bool, n)
for j := range dp[i] {
dp[i][j] = make([]bool, n+1)
}
}
for i := 0; i < n; i++ {
for j := 0; j < n; j++ {
dp[i][j][1] = s1[i] == s2[j]
}
}
for l := 2; l < n+1; l++ {
for i1 := 0; i1 < n-l+1; i1++ {
for i2 := 0; i2 < n-l+1; i2++ {
for i := 1; i < l; i++ {
if dp[i1][i2][i] && dp[i1+i][i2+i][l-i] {
dp[i1][i2][l] = true
break
}
if dp[i1][i2+l-i][i] && dp[i1+i][i2][l-i] {
dp[i1][i2][l] = true
break
}
}
}
}
}
return dp[0][0][n]
}


• function isScramble(s1: string, s2: string): boolean {
const n = s1.length;
const f = new Array(n)
.fill(0)
.map(() => new Array(n).fill(0).map(() => new Array(n + 1).fill(-1)));
const dfs = (i: number, j: number, k: number): boolean => {
if (f[i][j][k] !== -1) {
return f[i][j][k] === 1;
}
if (k === 1) {
return s1[i] === s2[j];
}
for (let h = 1; h < k; ++h) {
if (dfs(i, j, h) && dfs(i + h, j + h, k - h)) {
return Boolean((f[i][j][k] = 1));
}
if (dfs(i + h, j, k - h) && dfs(i, j + k - h, h)) {
return Boolean((f[i][j][k] = 1));
}
}
return Boolean((f[i][j][k] = 0));
};
return dfs(0, 0, n);
}