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Solution.java
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package ScrambleString;
import java.util.HashMap;
import java.util.Map;
/**
* User: Danyang
* Date: 1/28/2015
* Time: 19:06
*
* 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.
*/
public class Solution {
/**
* Algorithm: recursive
* treat as set of chars
* partition the string
*
* Notice:
* 1. prune, otherwise TLE
* @param s1
* @param s2
* @return
*/
public boolean isScramble(String s1, String s2) {
if(s1.equals(s2))
return true;
if(!relaxedEqual(s1, s2))
return false;
for(int l=1; l<s1.length(); l++)
if(isScramble(s1.substring(0, l), s2.substring(0, l)) &&
isScramble(s1.substring(l, s1.length()), s2.substring(l, s2.length())) ||
isScramble(s1.substring(0, l), s2.substring(s2.length()-l, s2.length())) &&
isScramble(s1.substring(l, s1.length()), s2.substring(0, s2.length()-l)))
return true;
return false;
}
boolean relaxedEqual(String s1, String s2) {
Map<Character, Integer> bucket = new HashMap<>();
for(char c: s1.toCharArray()) {
if(!bucket.containsKey(c))
bucket.put(c, 0);
bucket.put(c, bucket.get(c)+1);
}
for(char c: s2.toCharArray()) {
if(!bucket.containsKey(c) || bucket.get(c)<1)
return false;
bucket.put(c, bucket.get(c)-1);
}
for(Map.Entry<Character, Integer> e: bucket.entrySet()) {
if(e.getValue()!=0)
return false;
}
return true;
}
}