Recursive Remove Char

TIP101 Unit 7 Session 2 (Click for link to problem statements)

Problem Highlights

• 💡 Difficulty: Easy
• ⏰ Time to complete: 10 mins
• 🛠️ Topics: Recursion, String Manipulation

1: U-nderstand

Understand what the interviewer is asking for by using test cases and questions about the problem.

• Established a set (2-3) of test cases to verify their own solution later.
• Established a set (1-2) of edge cases to verify their solution handles complexities.
• Have fully understood the problem and have no clarifying questions.
• Have you verified any Time/Space Constraints for this problem?

• Q: What should the function return if the string is empty?
  • A: If the string is empty, the function should return an empty string.

HAPPY CASE
Input: s = "banana", char = "a"
Output: "bnn"
Explanation: All instances of 'a' are removed from the string.

EDGE CASE
Input: s = "aaaaa", char = "a"
Output: ""
Explanation: Removing 'a' from a string of only 'a's results in an empty string.

2: M-atch

Match what this problem looks like to known categories of problems, e.g. Linked List or Dynamic Programming, and strategies or patterns in those categories.

This problem is a straightforward application of recursive string manipulation:

• Removing characters from a string recursively by breaking down the problem into smaller segments.

3: P-lan

Plan the solution with appropriate visualizations and pseudocode.

General Idea: Use recursion to process each character of the string, removing the specified character and building the result recursively.

1) If the string is empty, return an empty string.
2) If the first character of the string is the character to remove, return the result of recursively processing the rest of the string (the first character is dropped).
3) Otherwise, prepend the first character to the result of recursively processing the rest of the string.

⚠️ Common Mistakes

• Failing to handle the base case where the string is empty, which could result in infinite recursion.
• Forgetting to prepend the first character in step 3, which silently drops characters that should be kept.

4: I-mplement

Implement the code to solve the algorithm.

def remove_char(s, char):
    if not s:
        return ""
    if s[0] == char:
        return remove_char(s[1:], char)
    else:
        return s[0] + remove_char(s[1:], char)

5: R-eview

Review the code by running specific example(s) and recording values (watchlist) of your code’s variables along the way.

• Test the function with input (“banana”, “a”) to ensure it returns “bnn”.
• Check the edge case (“aaaaa”, “a”) to confirm it returns an empty string.

6: E-valuate

Evaluate the performance of your algorithm and state any strong/weak or future potential work.

• Time Complexity: O(n^2) where n is the length of the string. There are n recursive calls, and each call computes s[1:], which allocates a new string in O(n) time per call (Python string slicing is O(k) for a slice of length k).
• Space Complexity: O(n^2) in the worst case (no characters removed). The recursion reaches depth n, and each frame retains a sliced suffix of sizes n, n-1, ..., 1, so the total memory held across the stack is O(n^2).