Check if a string can become empty by recursive deletion using Slicing

Given a string 'S' and a substring 'sub', the task is to determine whether 'S' can be reduced to an empty string by repeatedly deleting one occurrence at a time of the substring 'sub' whenever it appears in the string.

Example:

Input:
s = "aaaaaa"
sub = "aa"

Output:
True (string becomes empty)

Explanation:
"aaaaaa" → remove "aa" → "aaaa"
"aaaa" → remove "aa" → "aa"
"aa" → remove "aa" → ""

While-loop with slicing

This method keeps checking if the substring exists in the main string, and whenever it does, we remove it using slicing. The loop continues until no more removals are possible, and then we check if the string is empty.

False

Explanation:

• s.find(sub): Searches for the substring 'sub' inside 's' and returns its starting index.
• If ind == -1: The substring is not found anymore, so the loop stops.
• s[:ind] + s[ind + len(sub):]: Removes the found substring by keeping the part before it and the part after it.
• res = (s == ""): Checks whether the final string is empty.

Recursive Deletion Using Slicing

Recursive deletion using slicing means removing a substring from a string again and again by cutting it out using slicing (s[: ]). The process keeps calling itself until either the string becomes empty or no more deletions are possible.

False

Explanation:

• if not s: If the string is empty, return True.
• ind = s.find(sub): Find the first occurrence of the substring.
• If found: Remove it using slicing: s[:ind] + s[ind + len(sub):]
• Call function again with the updated string.
• If not found: Return False because the string can't be reduced further.

Using replace() Recursively

This method removes the substring using replace() inside a recursive function. Each recursive call removes the substring again and again until either the string becomes empty or no more removals are possible.

False

Explanation:

• def fun(s, sub): Defines a recursive function that keeps removing the substring.
• if not s: Base case: if the string becomes empty, return True.
• s1 = s.replace(sub, ""): Removes all occurrences of sub from s and stores the updated string in s1.
• if s1 == s: Checks if nothing was removed.
• If true: substring no longer exists return False.

Iterative replace() Loop

An iterative replace() loop repeatedly calls replace() inside a loop to update a string step-by-step until all desired changes are applied.

False

Explanation:

• prev = None: tracks previous state to detect changes.
• while s != prev: loop until 's' stops changing.
• prev = s: save current state before replacement.
• s = s.replace(sub, ""): remove all instances of 'sub'.

Related Articles:

• Check if a string consisting only of a, b, c can be made empty by removing substring "abc" recursively
• Count of ways to empty given String by recursively removing all adjacent duplicates