Условие
Decide whether text can be split into a sequence of words from words. Words may be reused. Dynamic programming over prefixes solves it efficiently.
Implement can_break(text, words).
Input
text: Lowercase letters only. words: The dictionary; each word has 1–50 lowercase letters.
Output
Return true if a split exists.
Ограничения
0 ≤ length of text ≤ 5,000; 1 ≤ words ≤ 1,000
Примеры
Пример 1
Ввод:
text = "applepenapple"
words = ["apple", "pen"]
Ожидаемый вывод:
expected = TrueВвод:
std::string text = "applepenapple";
std::vector<std::string> words = {"apple", "pen"};
Ожидаемый вывод:
bool expected = true;Ввод:
string text = "applepenapple";
string[] words = new string[] {"apple", "pen"};
Ожидаемый вывод:
bool expected = true;Пояснение
"apple" + "pen" + "apple", reusing "apple".
Пример 2
Ввод:
text = "catsandog"
words = ["cats", "dog", "sand", "and", "cat"]
Ожидаемый вывод:
expected = FalseВвод:
std::string text = "catsandog";
std::vector<std::string> words = {"cats", "dog", "sand", "and", "cat"};
Ожидаемый вывод:
bool expected = false;Ввод:
string text = "catsandog";
string[] words = new string[] {"cats", "dog", "sand", "and", "cat"};
Ожидаемый вывод:
bool expected = false;Пояснение
Every split leaves "og" or "ando" unmatched: false.
Ссылки
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