Date: 2026-06-06
Time: 17:30
make-the-string-great/solution.pyThis file solves LeetCode 1544 — Make The String Great. The problem: given a string of mixed-case English letters, repeatedly remove any adjacent pair where one character is the uppercase version of the other (e.g., "aA" or "Aa"), until no such pair remains.
Note: the method is misnamed goodNodes — it should be makeGood per the LeetCode signature. The docstring is correct about what it does.
Solution.goodNodes(self, s: str) -> str — the single entry point. Takes a string, returns the "great" version with all bad adjacent pairs removed.
Stack-based pair cancellation. This is the canonical pattern for problems where adjacent elements can annihilate each other (see also: remove-all-adjacent-duplicates-in-string). Instead of repeatedly scanning the string for pairs (which would be O(n²)), the stack processes each character once:
The trick for detecting a bad pair is abs(ord(stack[-1]) - ord(c)) == 32. This works because in ASCII, every lowercase letter is exactly 32 higher than its uppercase counterpart (ord('a') - ord('A') == 32). Two characters that are the same letter in different cases will always differ by exactly 32. Two characters that are different letters — even across case — will never differ by exactly 32, because no two distinct letters occupy adjacent slots 32 apart in the ASCII table.
Imports: None — pure standard library, no external dependencies.
Imported by: The "Imported By" list in the prompt is misleading — those ~400+ test files are importing from their own local solution.py, not from this file. The actual consumer is make-the-string-great/test_solution.py.
1. Initialize an empty stack (list).
2. Iterate through each character c in s.
3. For each character, check if the stack is non-empty and the top element forms a case-inverse pair with c (ASCII difference of 32).
4. If yes: pop the stack — both characters are eliminated.
5. If no: push c onto the stack.
6. Join the stack into a string and return.
The stack naturally handles chain reactions. When a pop exposes a new top, the *next* incoming character is checked against that new top — so cascading cancellations like "abBAc" → "aAc" → "c" are handled in a single pass without re-scanning.
|upper - lower| == 32 for every English letter. This is always true for standard Python str.s is empty, the loop doesn't execute and "".join([]) returns "".None. The function trusts its input — no validation that s contains only English letters. Non-alphabetic characters would pass through the stack unpaired, which is correct behavior for the problem constraints.