Update q1.py

[Kv22a]

No description provided.

[gemini-code-assist]

Summary of Changes

Hello @Kv22a, I'm Gemini Code Assist¹! I'm currently reviewing this pull request and will post my feedback shortly. In the meantime, here's a summary to help you and other reviewers quickly get up to speed!

This pull request addresses the TODO in q1.py by providing a complete implementation for the first_stable_character function. The function's purpose is to locate and return a specific type of 'stable' character within an input string, enhancing the utility of the q1.py module.

Highlights

• q1.py function implementation: The first_stable_character function has been implemented in q1.py. This function identifies the first character that appears at least twice consecutively and does not appear again later in the string.

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[gemini-code-assist]

Code Review

This pull request implements the logic for the first_stable_character function. The core logic correctly solves the problem as described. My review provides a single, comprehensive suggestion to improve the code's readability and adherence to Python best practices by removing redundant code, cleaning up a leftover comment, and making the function's exit points more explicit.

[gemini-code-assist]

The implementation is logically correct, but it can be made cleaner and more idiomatic.

• The pass statements on lines 56 and 67 are unnecessary and can be removed.
• The TODO comment on line 66 should be removed now that the solution is implemented.
• Adding an explicit return None at the end of the function makes the intent clearer than relying on an implicit return when no stable character is found.

Here is a suggested revision that incorporates these improvements.

Suggested change

	n = len(s)
	i = 0
	pass
	while i < n:
	char = s[i]
	start = i
	while i < n and s[i] == char:
	i += 1
	end = i
	if end - start >= 2:
	if char not in s[end:]:
	return char # TODO: Implement your solution here
	pass
	n = len(s)
	i = 0
	while i < n:
	char = s[i]
	start = i
	while i < n and s[i] == char:
	i += 1
	end = i
	if end - start >= 2:
	if char not in s[end:]:
	return char
	return None