The String Interleaving Problem is a fascinating problem that is often encountered in algorithm interviews and dynamic programming discussions. At its core, the problem tests your understanding of how strings can be manipulated and combined, while adhering to specific rules.

Problem Definition

Given three strings: s1, s2, and s3, we need to determine if s3 is formed by interleaving s1 and s2. This means that all characters from s1 and s2 must be present in s3, and the relative order of characters must be preserved.

For example, if we have:

• s1 = "abc"
• s2 = "def"
• s3 = "adbcef"

In this case, s3 is indeed an interleaving of s1 and s2: the character 'a' comes from s1, 'd' from s2, 'b' from s1, 'c' from s1, 'e' from s2, and 'f' from s2.

Conversely, if we had:

• s1 = "abc"
• s2 = "def"
• s3 = "abdecf"

Here, s3 is not an interleaving of s1 and s2 because 'd' precedes 'e' unexpectedly, violating the order requirement.

Dynamic Programming Approach

To solve the String Interleaving Problem using dynamic programming (DP), we can create a 2D table where each cell dp[i][j] indicates whether the first i characters of s1 and the first j characters of s2 can form the first i+j characters of s3.

Here's how you can visualize this:

1. Initialization:

   • Create a DP table with dimensions (len(s1) + 1) x (len(s2) + 1).
   • Set dp[0][0] to True since an empty s1 and s2 can produce an empty s3`.

2. Filling the Table:

   • Iterate through each character of s1 and s2, and fill in dp as follows:
     • If the current character of s1 matches the corresponding character in s3, carry over the value from the cell to the left (dp[i-1][j]).
     • If the current character of s2 matches the corresponding character in s3, carry over the value from the cell above (dp[i][j-1]).
   • The final value at dp[len(s1)][len(s2)] will indicate if s3 can be formed by interleaving s1 and s2.

Example Code

Here’s a basic implementation in Python:

def is_interleave(s1: str, s2: str, s3: str) -> bool:
    if len(s1) + len(s2) != len(s3):
        return False

    dp = [[False] * (len(s2) + 1) for _ in range(len(s1) + 1)]
    dp[0][0] = True

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

    for j in range(1, len(s2) + 1):
        dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]

    for i in range(1, len(s1) + 1):
        for j in range(1, len(s2) + 1):
            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[len(s1)][len(s2)]

Explanation of the Code

• Edge Case: Initially, we check if the lengths of s1, s2, and s3 are compatible. If not, we return False.
• DP Table Creation: A table dp is constructed. The dimensions depend on the lengths of s1 and s2.
• Filling Initial Rows and Columns: The first row and first column are filled based on whether characters in s1 and s2 match with s3.
• Nested Loops: We then fill in each cell of the DP table based on previous results.

Time and Space Complexity

The time and space complexity for this algorithm is O(n * m), where n is the length of s1 and m is the length of s2. This is because we are filling out a 2D table with dimensions related to these strings.

Conclusion

The String Interleaving Problem is a critical problem in dynamic programming, intricately tied to string manipulations. By understanding how to build the DP table and trace through the strings, we can tackle this problem efficiently. As you practice more dynamic programming problems, solutions like this will become second nature, enhancing your problem-solving skill set significantly.