Date: 2026-06-06
Time: 15:54
count-asterisks/solution.pyThis file solves LeetCode 2315: Count Asterisks. It owns the single responsibility of counting * characters in a string while ignoring any * that falls between a matched pair of | delimiters. It's the canonical solution module for this problem in the leetcode-implementations repo.
countstarsexceptbetweenpair(s: str) -> int — The sole function. Its contract:
s containing lowercase letters, |, and *. The problem guarantees an even number of | characters, so every | has a partner.* characters that are *not* enclosed between any paired | delimiters.Toggle-flag state machine — The function uses a boolean inside as a two-state automaton. Each | flips the state. This is the standard idiom for "skip content between matched delimiters" when delimiters don't nest and always come in pairs. It avoids the overhead of splitting or regex and processes the string in a single pass.
The approach is O(n) time, O(1) space — no auxiliary data structures, no string slicing.
Imports: None. The function uses only Python builtins.
Imported by: The "Imported By" list in the prompt is misleading — those hundreds of test files are unrelated problems. The actual consumer is count-asterisks/test_solution.py. The other test files likely share a common test harness or conftest that imports broadly, not this function specifically.
1. Initialize count = 0 (accumulator) and inside = False (state flag).
2. Iterate character-by-character over s.
3. On |: toggle inside. This pairs the 1st | with the 2nd, the 3rd with the 4th, etc.
4. On * when inside is False: increment count. Asterisks inside paired bars are silently skipped.
5. All other characters (lowercase letters, * when inside): ignored.
6. Return the final count.
For input "l|*e*et|c**o|*de|":
l → outside, not *, skip| → toggle to inside*e*et → inside, all skipped| → toggle to outsidec**o → outside, two * counted → count = 2| → toggle to inside*de → inside, skipped| → toggle to outside| count is a precondition from the problem statement. If violated, the function still runs without error but inside would be True at the end, which could silently miscount trailing asterisks. The function does not validate this.| form a pair, the 3rd and 4th form a pair, etc. There's no nesting or matching logic — the toggle naturally enforces left-to-right sequential pairing.None. The function assumes valid input per the problem constraints. An empty string returns 0 (the loop simply doesn't execute). No exceptions are raised or caught.
count-asterisks/test_solution.py — See what edge cases are tested (empty string, no bars, consecutive bars, all-asterisk segments)count-asterisks/plan.md — The problem analysis and approach reasoning before implementationmaximum-nesting-depth-of-the-parentheses/solution.py:maxDepth — A related toggle/counter pattern but tracking depth instead of a binary flagtoggle-flag-vs-split-approach — Compare this single-pass toggle to the alternative of s.split('|') and summing * counts at even indicescount-asterisks-linear-scan — countstarsexceptbetweenpair processes the input in a single O(n) pass with O(1) auxiliary spacecount-asterisks-toggle-pairing — Pipe characters are paired sequentially by position (1st with 2nd, 3rd with 4th) via a boolean toggle, not by any matching or nesting logiccount-asterisks-no-validation — The function does not validate that | count is even; it trusts the caller to satisfy this preconditioncount-asterisks-pure-function — The function is pure with no side effects, no imports, and no mutable state beyond local variables