lc 87 scramble string

LC 87 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.

Example 1:

Input: s1 = “great”, s2 = “rgeat”
Output: true
Example 2:

Input: s1 = “abcde”, s2 = “caebd”

Output: false

Tag: Enumeration, Recursion

The important idea is that if the number of characters is equal between the string1 and string2, that means string2 is one of string1’s enumerations. And we all know that one enumeration can be transform to another one by swapping.

So, what we need to do is just check whether the two strings are enumerations.

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class Solution {
public:
bool isScramble(string s1, string s2) {
if(s1 == s2)
return true
vector<char> alpha(26, 0)
for(int i = 0; i < s1.length(); i++){
alpha[s1[i] - 'a']++
alpha[s2[i] - 'a']--
}
for(int i = 0; i < 26; i++){
if(alpha[i] != 0)
return false
}
for(int i = 1; i < s1.length(); i++){
if(isScramble(s1.substr(0, i), s2.substr(0, i)) && isScramble(s1.substr(i), s2.substr(i)))
return true
if(isScramble(s1.substr(0, i), s2.substr(s1.length()-i)) && isScramble(s1.substr(i), s2.substr(0, s1.length()-i)))
return true
}
return false
}
}