Approach

To effectively answer the question about writing a function that checks if a given string is an interleaving of two other strings, follow this structured framework:

1. Understand the Problem: Clearly define what it means for a string to be an interleaving of two other strings.

2. Identify Inputs and Outputs: Specify the strings involved and what the function should return.

3. Choose an Appropriate Algorithm: Consider dynamic programming or recursion to solve the problem.

4. Implement the Solution: Write the function in a clear and efficient manner.

5. Test the Function: Include test cases to validate the solution.

Key Points

• Definition of Interleaving: A string s3 is an interleaving of s1 and s2 if it can be formed by merging s1 and s2 such that the order of characters in s1 and s2 is preserved.

• Inputs and Outputs: The function should take three strings (s1, s2, and s3) and return a boolean indicating if s3 is an interleaving of s1 and s2.

• Algorithm Choice: Dynamic programming is often preferred for its efficiency, especially for longer strings.

• Edge Cases: Consider cases where lengths of s1, s2, and s3 do not match.

Standard Response

Here’s a sample implementation of a function that checks if s3 is an interleaving of s1 and s2:

def is_interleaving(s1: str, s2: str, s3: str) -> bool:
 # First, check the lengths of the strings
 if len(s1) + len(s2) != len(s3):
 return False
 
 # Create a DP table
 dp = [[False] * (len(s2) + 1) for _ in range(len(s1) + 1)]
 
 # Initialize the DP table
 dp[0][0] = True
 
 # Fill in the first row
 for j in range(1, len(s2) + 1):
 dp[0][j] = dp[0][j - 1] and s2[j - 1] == s3[j - 1]
 
 # Fill in the first column
 for i in range(1, len(s1) + 1):
 dp[i][0] = dp[i - 1][0] and s1[i - 1] == s3[i - 1]
 
 # Fill in the rest of the DP table
 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)]

Tips & Variations

Common Mistakes to Avoid

• Ignoring Lengths: Failing to check if the combined lengths of s1 and s2 equal s3 can lead to unnecessary computation.

• Not Using Dynamic Programming: Recursion without memoization can lead to excessive time complexity.

• Overlooking Edge Cases: Neglecting single-character comparisons or completely empty strings can lead to incorrect results.

Alternative Ways to Answer

• Recursive Approach: Some might prefer a simpler recursive check, though it may not be as efficient for larger strings.

Role-Specific Variations

• Technical Positions: Emphasize time and space complexity analysis of the solution.

-