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Dzhoker1
reviewed
Nov 25, 2018
| hashesarray = [] | ||
| for i in range(0, len(s)): | ||
| for j in range(0, len(s)): | ||
| if hash(s[i:j + 1]) not in hashesarray: |
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Каждая проверка наличия хэша в списке имеет сложность O(n), отсюда и такой разрыв во времени выполнения.
| def mod_count_substrings(s): | ||
| d = collections.OrderedDict() | ||
| for i in range(0, len(s)): | ||
| for j in range(0, len(s)): |
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Второй цикл можно начинать не от 0, а от j.
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| # Проверяем работу функции | ||
| print(count_substrings('mama')) | ||
| print(count_substrings('mama')) |
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