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.
Solution
- Recursion + pruning.
- 3-dimensional dp.
dp[k][i][j] == true
means string s1(start from i, length k) is a scrambled string of string s2(start from j, length k).
Code
public class Solution {
public boolean isScramble_1(String s1, String s2) {
if (s1.compareTo(s2) == 0) return true;
if (s1.length() != s2.length()) return false;
return isScrambleRe(s1, s2);
}
public boolean isScrambleRe(String s1, String s2) {
if (hasSameLetters(s1, s2) == false) return false;
int len = s1.length();
if (len == 0 || len == 1) return true;
for (int i = 1; i < len; ++i) {
if (isScrambleRe(s1.substring(0,i), s2.substring(0,i))
&& isScrambleRe(s1.substring(i), s2.substring(i))
|| isScrambleRe(s1.substring(0,i), s2.substring(len-i))
&& isScrambleRe(s1.substring(i), s2.substring(0,len-i))) {
return true;
}
}
return false;
}
public boolean hasSameLetters(String s1, String s2) {
if (s1.compareTo(s2) == 0) return true;
if (s1.length() != s2.length()) return false;
int[] count = new int[256];
for (int i = 0; i < s1.length(); ++i) count[(int)s1.charAt(i)]++;
for (int i = 0; i < s2.length(); ++i) count[(int)s2.charAt(i)]--;
for (int i = 0; i < 256; ++i)
if (count[i] != 0) return false;
return true;
}
public boolean isScramble(String s1, String s2) {
if (s1.compareTo(s2) == 0) return true;
if (s1.length() != s2.length()) return false;
int N = s1.length();
boolean[][][] dp = new boolean[N+1][N][N];
for (int k = 1; k <= N; ++k) {
for (int i = 0; i <= N - k; ++i) {
for (int j = 0; j <= N - k; ++j) {
dp[k][i][j] = false;
if (k == 1) dp[1][i][j] = (s1.charAt(i) == s2.charAt(j));
for (int p = 1; p < k && !dp[k][i][j]; ++p) {
if (dp[p][i][j] && dp[k-p][i+p][j+p] || dp[p][i][j+k-p] && dp[k-p][i+p][j])
dp[k][i][j] = true;
}
}
}
}
return dp[N][0][0];
}
}