String interleaving is a fascinating concept that often appears in algorithmic coding interviews, particularly those focused on string manipulation. At its core, string interleaving checks whether a string can be formed by interleaving two other strings while maintaining the original order of characters from the two source strings. Let’s break this down step by step.

What is String Interleaving?

Suppose we have two strings, A and B, along with a third string C. The task is to determine if C can be formed by interleaving characters from A and B. Interleaving means that you can take characters from A and B and combine them in a way that preserves the sequence of characters in both strings.

For example:

• A = "abc"
• B = "def"
• C = "adbcef" (This is a valid interleaving as we can derive C from A and B)

However, not all combinations will work. Consider:

• C = "abdecf" (This is not valid because the 'd' from B comes after 'e', which violates the sequence defined in B)

The Problem Breakdown

When facing a problem of string interleaving, you can approach it in several ways. Below, we will cover a more structured approach using dynamic programming, which is often preferred in coding interviews due to its efficiency.

Dynamic Programming Approach

The dynamic programming (DP) solution involves creating a 2D table (or matrix) to store boolean values that indicate whether a substring of C can be formed by interleaving substrings of A and B. The basic idea is to iterate through the lengths of A and B, checking character by character.

Here’s a step-by-step guide to constructing the algorithm:

1. Initialize a DP Table: Create a 2D array dp[i][j] where dp[i][j] will be true if the first i characters of A and the first j characters of B can form the first i + j characters of C.

2. Base Case: Set dp[0][0] to true since two empty strings can form an empty string.

3. Filling the DP Table:

   • Iterate through each index of A and B.
   • For each entry dp[i][j], check:
     • If A[i-1] equals C[i+j-1], carry over the truth value from dp[i-1][j].
     • If B[j-1] equals C[i+j-1], carry over the truth value from dp[i][j-1].

Here’s what the code looks like in Python:

def isInterleave(A, B, C):
    if len(A) + len(B) != len(C):
        return False

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

    for i in range(len(A) + 1):
        for j in range(len(B) + 1):
            if i > 0 and A[i - 1] == C[i + j - 1]:
                dp[i][j] = dp[i][j] or dp[i - 1][j]
            if j > 0 and B[j - 1] == C[i + j - 1]:
                dp[i][j] = dp[i][j] or dp[i][j - 1]

    return dp[len(A)][len(B)]

Example Walkthrough

Let’s examine how the algorithm works using the example where:

• A = "aab"
• B = "axy"
• C = "aaxyab"

1. Initialization: We verify that the lengths match (A + B = C).
2. DP Table Fill: Starting from the base case, we fill in our DP table by checking character matches:
   • dp[0][0] = True (both strings empty)
   • Then we move on to check indices and populate based on the rules we established.

Eventually, if we complete this table, we can inspect dp[len(A)][len(B)] to see whether we can interleave A and B to create C.

Time Complexity

The time complexity of this approach is O(n * m), where n is the length of string A and m is the length of string B. The space complexity can also be optimized to O(min(n, m)) if needed, as only the previous row is required for further calculations.

With this understanding of string interleaving, you're better equipped to tackle related problems that may come up in interviews. Practicing variations of these problems will enhance your string manipulation skills and overall problem-solving abilities in coding contests and assessments.