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.
class Solution {
public:
bool check(string s1, string s2) {
if(s1.length() != s2.length()) return false;
string cp1 = s1, cp2 = s2;
sort(cp1.begin(), cp1.end());
sort(cp2.begin(), cp2.end());
for(int i=; i<cp1.length(); ++i) {
if(cp1[i] != cp2[i]) return false;
}
return true;
}
bool dfs(string s1, string s2) {
int m = s1.length(), n = s2.length();
if(!check(s1, s2)) return false;
if(m == ) {
if(s1 == s2) return true;
return false;
}
string l, r, p, q;
for(int le = ; le < m; ++le) {
l = s1.substr(, le);
r = s1.substr(le, m - le);
p = s2.substr(, le);
q = s2.substr(le, m - le);
if(dfs(l, p) && dfs(r, q)) return true;
else {
p = s2.substr(m - le, le);
q = s2.substr(, m - le);
if(dfs(l, p) && dfs(r, q)) return true;
}
}
return false;
}
bool isScramble(string s1, string s2) {
int m = s1.length(), n = s2.length();
if(m != n) return false;
return dfs(s1, s2);
}
};