Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.
An interleaving of two strings s and t is a configuration where they are divided into non-empty substrings such that:
-
s = s1 + s2 + ... + sn -
t = t1 + t2 + ... + tm -
|n - m| <= 1 - The interleaving is
s1 + t1 + s2 + t2 + s3 + t3 + ...ort1 + s1 + t2 + s2 + t3 + s3 + ...
Note: a + b is the concatenation of strings a and b.
Example 1:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbcbcac"
Output: true
Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc"
Output: false
Example 3:
Input: s1 = "", s2 = "", s3 = ""
Output: true
Constraints:
-
0 <= s1.length, s2.length <= 100 -
0 <= s3.length <= 200 -
s1,s2, ands3consist of lowercase English letters.
Follow up: Could you solve it using only O(s2.length) additional memory space?
SOLUTION:
class Solution:
def isLeave(self, s1, s2, s3, i, j, k, m, n, p):
if (i, j, k) in self.cache:
return self.cache[(i,j,k)]
if i >= m or j >= n or k >= p:
if i >= m:
res = s2[j:] == s3[k:]
self.cache[(i,j,k)] = res
return res
if j >= n:
res = s1[i:] == s3[k:]
self.cache[(i,j,k)] = res
return res
self.cache[(i,j,k)] = False
return False
if s1[i] == s3[k] and self.isLeave(s1, s2, s3, i + 1, j, k + 1, m, n, p):
self.cache[(i,j,k)] = True
return True
if s2[j] == s3[k] and self.isLeave(s1, s2, s3, i, j + 1, k + 1, m, n, p):
self.cache[(i,j,k)] = True
return True
self.cache[(i,j,k)] = False
return False
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
m = len(s1)
n = len(s2)
p = len(s3)
self.cache = {}
return self.isLeave(s1, s2, s3, 0, 0, 0, m, n, p)
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