Description
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 s
and t
are divided into n
and m
substrings respectively, 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 Explanation: One way to obtain s3 is: Split s1 into s1 = "aa" + "bc" + "c", and s2 into s2 = "dbbc" + "a". Interleaving the two splits, we get "aa" + "dbbc" + "bc" + "a" + "c" = "aadbbcbcac". Since s3 can be obtained by interleaving s1 and s2, we return true.
Example 2:
Input: s1 = "aabcc", s2 = "dbbca", s3 = "aadbbbaccc" Output: false Explanation: Notice how it is impossible to interleave s2 with any other string to obtain s3.
Example 3:
Input: s1 = "", s2 = "", s3 = "" Output: true
Constraints:
0 <= s1.length, s2.length <= 100
0 <= s3.length <= 200
s1
,s2
, ands3
consist of lowercase English letters.
Follow up: Could you solve it using only O(s2.length)
additional memory space?
Solution
Python3
class Solution:
def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
M, N, K = len(s1), len(s2), len(s3)
if M + N != K: return False
@cache
def go(i, j):
if i + j == K: return True
if i < M and s1[i] == s3[i + j] and go(i + 1, j):
return True
if j < N and s2[j] == s3[i + j] and go(i, j + 1):
return True
return False
return go(0, 0)