問題
For each requested pattern, decide whether it appears as a contiguous substring anywhere in the shared text. Patterns may overlap and may repeat. Return results in the original query order. The limits require an approach that can handle many patterns efficiently.
Implement find_many_patterns(text, patterns).
Input
text: The text searched by every query.
patterns: The patterns, in query order.
Output
Return YES for each occurring pattern and NO for each absent one.
制約
1…100,000 text characters. 1…100,000 nonempty patterns, with total length at most 500,000. All characters are lowercase a–z.
例
入力：
text = "bananas"
patterns = ["ana", "apple", "nas"]
期待される出力：
expected = ["YES", "NO", "YES"]入力：
std::string text = "bananas";
std::vector<std::string> patterns = {"ana", "apple", "nas"};
期待される出力：
std::vector<std::string> expected = {"YES", "NO", "YES"};入力：
string text = "bananas";
string[] patterns = new string[] {"ana", "apple", "nas"};
期待される出力：
string[] expected = new string[] {"YES", "NO", "YES"};解説
Ana appears starting at position 2 and nas at position 5 (positions start at 1 here). Apple never appears.
入力：
text = "aaaa"
patterns = ["aa", "aaaaa", "aa"]
期待される出力：
expected = ["YES", "NO", "YES"]入力：
std::string text = "aaaa";
std::vector<std::string> patterns = {"aa", "aaaaa", "aa"};
期待される出力：
std::vector<std::string> expected = {"YES", "NO", "YES"};入力：
string text = "aaaa";
string[] patterns = new string[] {"aa", "aaaaa", "aa"};
期待される出力：
string[] expected = new string[] {"YES", "NO", "YES"};解説
Aa occurs despite overlapping matches. Five a characters cannot fit in a four-character text. The repeated aa query receives the same answer.
解き方のヒント
まずは自分で考えてから、ヒントを 1 つずつ開いてみましょう。
ヒント 1
複数のパターンで作業を共有できるので、各クエリごとに別々に本文を探さないようにする。
ヒント 2
全パターンを入れた1つの構造を作り、本文は1回だけ走査して、文字が来るたびに一致を進める。
ヒント 3
重複パターンはクエリ順に報告し、重なり合う一致もそれぞれ独立に数える。
目安の計算量
時間 O(text length + total pattern length)、追加メモリ O(total pattern length)
参考資料
この課題はCodeUtility用に調整されています。上記の説明と制約に従ってください。