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Mô-đun 2 · Bài học 4 trong tổng số 48/32 trong khóa học~12 min
Bài học theo mô-đun (4/4)

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When a greedy quantifier "overshoots", the engine performs backtracking: it steps back, tries another configuration, retries. On simple patterns it is invisible; on poorly written patterns it can become exponential and freeze the browser for seconds. It is the root of ReDoS (Regular expression Denial of Service).

Code
Pattern: .*\d
Sample:  abc123def
Steps:
  1.  .*  consumes everything: "abc123def"
  2.  \d  fails (end of string).
  3.  Backtrack: .* drops 'f' -> "abc123de", \d fails on 'f'.
  4.  Backtrack: .* drops 'ef' -> "abc123d", \d fails on 'e'.
  5.  ... and so on until \d finds a digit: match = "abc123" + "?"... actually
       the pattern matches "abc123" because .* leaves "def" and \d looks at '3'? No: the
       backtrack continues and .* ends up being "abc12", \d=3.  Final match: "abc123".

The engine always gets there, but if a greedy quantifier contains another greedy quantifier (e.g. (a+)+), the number of configurations to try can explode.

Catastrophic patterns

Code
Pattern: (a+)+b
Sample:  aaaaaaaaaaaaaaaaaaaaa

No trailing b → the pattern fails, but to get there the engine tries every possible partition of the as. Exponential growth: with 30 as you can wait for hours.

Perils of backtracking and ReDoS

When a pattern contains overlapping or nested quantifiers (e.g. (a+)+), the engine evaluates an exponential number of paths before failing. Replacing generic wildcards with excluding character classes dramatically reduces backtracking paths.

Try it

tập thể dục#regex.m2.l4.e1
Nỗ lực: 0Đang tải…

Extract the content between square brackets for every occurrence in the text. Use the lazy version of the quantifier to avoid jumping to later closings.

Đang tải trình chỉnh sửa…
Hiển thị gợi ý

\\[.*?\\] stops at the first closing ]: no heavy backtracking.

Giải pháp khả dụng sau 3 lần thử

Review exercise

tập thể dục#regex.m2.l4.e2
Nỗ lực: 0Đang tải…

Find every `http://...` or `https://...` URL in the text, using the `?` quantifier for the optional `s` and `\\S+` for the non-space characters of the path.

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Hiển thị gợi ý

https? means 'http' followed by an optional 's'. Then //, then \\S+ (non-spaces).

Giải pháp khả dụng sau 3 lần thử

Additional challenge

tập thể dục#regex.m2.l4.e3
Nỗ lực: 0Đang tải…

Match strings enclosed in double quotes, avoiding excessive backtracking by using the negated class `[^"]*` instead of `.*`.

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Hiển thị gợi ý

Replace the dot with the negated class [^"] to restrict the engine's backtracking paths.

Giải pháp khả dụng sau 3 lần thử