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)

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Abhishek Chaudhary

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