Introduction

In this blog post, we will explore how to solve the LeetCode problem "97" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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?

Explanation

Here's the solution to the interleaving string problem, addressing efficiency and memory constraints:

• Dynamic Programming: The core idea is to use dynamic programming to build a table dp where dp[i][j] represents whether s3[0...i+j-1] is an interleaving of s1[0...i-1] and s2[0...j-1].
• Base Case and Transitions: The base case is dp[0][0] = True (empty strings interleave to form an empty string). The transitions involve checking if the current character of s3 matches either the current character of s1 or s2, and if so, inheriting the interleaving status from the previous state.

• Space Optimization: The DP table is optimized to use only O(s2.length) space. We only keep track of the previous row of the dp table as we iterate.

• Runtime Complexity: O(m*n), where n = len(s1) and m = len(s2).

• Storage Complexity: O(m), where m = len(s2).

Code

    def isInterleave(s1: str, s2: str, s3: str) -> bool:
    """
    Given strings s1, s2, and s3, find whether s3 is formed by an interleaving of s1 and s2.
    """

    n = len(s1)
    m = len(s2)
    l = len(s3)

    if n + m != l:
        return False

    dp = [False] * (m + 1)
    dp[0] = True

    for j in range(1, m + 1):
        dp[j] = dp[j - 1] and s2[j - 1] == s3[j - 1]

    for i in range(1, n + 1):
        dp[0] = dp[0] and s1[i - 1] == s3[i - 1]

        for j in range(1, m + 1):
            dp[j] = (dp[j] and s1[i - 1] == s3[i + j - 1]) or \
                    (dp[j - 1] and s2[j - 1] == s3[i + j - 1])

    return dp[m]