## DEV Community

Abhishek Chaudhary

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# Interleaving String

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 + ...` or `t1 + 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`, and `s3` consist 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)
``````