InterleavingString.py

"""
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 + ... 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
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, and s3 consist of lowercase English letters.


Follow up: Could you solve it using only O(s2.length) additional memory space?
"""


class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        len1, len2, len3 = len(s1), len(s2), len(s3)

        # If the lengths of s1 and s2 don't add up to the length of s3, return False
        if len1 + len2 != len3:
            return False

        # Create a 2D array dp to store intermediate results
        dp = [[False] * (len2 + 1) for _ in range(len1 + 1)]

        for i in range(len1 + 1):
            for j in range(len2 + 1):
                if i == 0 and j == 0:
                    dp[i][j] = True
                elif i == 0:
                    dp[i][j] = dp[i][j - 1] and s2[j - 1] == s3[i + j - 1]
                elif j == 0:
                    dp[i][j] = dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]
                else:
                    dp[i][j] = (dp[i - 1][j] and s1[i - 1] == s3[i + j - 1]) or (dp[i][j - 1] and s2[j - 1] == s3[i + j - 1])

        return dp[len1][len2]


# Test code
solution = Solution()

# Test case 1
s1 = "aabcc"
s2 = "dbbca"
s3 = "aadbbcbcac"
result1 = solution.isInterleave(s1, s2, s3)
print("Test case 1 result:", result1)  # Expected output: True

# Test case 2
s1 = "aabcc"
s2 = "dbbca"
s3 = "aadbbbaccc"
result2 = solution.isInterleave(s1, s2, s3)
print("Test case 2 result:", result2)  # Expected output: False

# Test case 3
s1 = ""
s2 = ""
s3 = ""
result3 = solution.isInterleave(s1, s2, s3)
print("Test case 3 result:", result3)  # Expected output: True