LeetCode in Python

97. Interleaving String

Medium

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

from typing import List, Optional

class Solution:
    def isInterleave(self, s1: str, s2: str, s3: str) -> bool:
        if len(s3) != (len(s1) + len(s2)):
            return False
        cache: List[List[Optional[bool]]] = [[None for _ in range(len(s2) + 1)] for _ in range(len(s1) + 1)]
        return self._isInterleave(s1, s2, s3, 0, 0, 0, cache)

    def _isInterleave(self, s1: str, s2: str, s3: str, i1: int, i2: int, i3: int, cache: List[List[Optional[bool]]]) -> bool:
        if cache[i1][i2] is not None:
            return cache[i1][i2]
        if i1 == len(s1) and i2 == len(s2) and i3 == len(s3):
            return True
        result = False
        if i1 < len(s1) and s1[i1] == s3[i3]:
            result = self._isInterleave(s1, s2, s3, i1 + 1, i2, i3 + 1, cache)
        if i2 < len(s2) and s2[i2] == s3[i3]:
            result = result or self._isInterleave(s1, s2, s3, i1, i2 + 1, i3 + 1, cache)
        cache[i1][i2] = result
        return result

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