# Solving Leetcode Interviews in Seconds with AI: Interleaving String


	# 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](https://www.chatmagic.app), 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
	```python
	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]
	```