Scramble String

Desicription

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":

1
2
3
4
5
6
7
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".

1
2
3
4
5
6
7
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".

1
2
3
4
5
6
7
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
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1 == s2)
return 1;
vector<int> cnt(26, 0);
for(int i = 0; s1[i]; i++)
cnt[s1[i] - 'a']++, cnt[s2[i] - 'a']--;
for(int i = 0; i < 26; i++)
if(cnt[i])
return 0;
for(int i = 1; s1[i]; i++){
if(isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
return 1;
if(isScramble(s1.substr(0, i), s2.substr(s1.size() - i)) && isScramble(s1.substr(i), s2.substr(0, s1.size() - i)))
return 1;
}
return 0;
}
};