Scramble String
Created: March 15, 2020 by [lek-tin]
Last updated: March 15, 2020
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
Solution (DFS with memoization)
from collections import Counter
class Solution:
def isScramble(self, s1: str, s2: str) -> bool:
if len(s1) == len(s2) == 0:
return True
return self.helper(s1, s2, {})
def helper(self, s1, s2, cache):
if( s1 == s2 ):
return True
if len(s1) <= 1:
return False
key = s1 + " " + s2
if key in cache:
return cache[key]
flag = False
N = len(s1)
for i in range(1, N):
if self.helper(s1[:i], s2[:i], cache) and self.helper(s1[i:], s2[i:], cache):
flag = True
break
if self.helper(s1[:i], s2[-i:], cache) and self.helper(s1[i:], s2[:-i], cache):
flag = True
break
cache[key] = flag
return flag
Solution