# leetcode: Scramble String

#### Scramble String Problem Description:

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*.

####
Scramble String Solution and **Precautions**:

(1) DP： F(i, j, len) denotes whether s1[i]…s1[i+len-1] and s2[j][j+len-1] are scramble string or not then we have the following recursive defintion:

F(i, j, len) = Union { F(i, j, l) and F(i + l, j + l, len – l) } OR

{ F(i + len – l, j, l) and F(i, j + l, len – l) }

for l = 1, … len – 1

OR

s1[i]…s1[i+len-1] == s2[j][j+len-1]

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This issue an O(N^3) solution which could pass OJ

(2) DFS + Pruning, essentially, we adopt the same recursive definition as in (1) and do dfs, once we find a possible Scramble string match, we return. This might be faster than DP since we don’t continue once we find possible Scramble string, but DP will anyways exhaust all the possible candidates.