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 isScramble(string s1, string s2) { // Note: The Solution object is instantiated only once and is reused by each test case. if(s1.length() !=s2.length() ) return false; //scramble s1 to s2. Partition s1 from 0 to n if the first n elements are the same in s1 and s2 except the order //keep a table to match the number of alphebits int A[26]={0}; for(int i=0;i<s1.length();i++) { A[s1[i]-'a']++; A[s2[i]-'a']--; } for(int i=0;i<s1.length();i++) if(A[s1[i]-'a'] != 0) return false; if(s1.length()==1) return true; //start partitioning bool result=false; for(int i=1;i<s2.length();i++) { bool result1=isScramble(s1.substr(0,i),s2.substr(0,i))&&isScramble(s1.substr(i,s1.length()-i),s2.substr(i,s2.length()-i)); bool result2=isScramble(s1.substr(0,i),s2.substr(s2.length()-i,i)) && isScramble(s1.substr(i,s1.length()-i),s2.substr(0,s2.length()-i)); result=result1||result2; if(result==true) return result; } return result; } };