Angele Z. #27
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Angele Z. #27
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CheezItMan
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Oct 9, 2019
CheezItMan
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CheezItMan
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Well done. You hit the learning goals here. Take a look at my comments and let me know if you have questions.
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| right = my_sentence.length - 1 | ||
| until left == right || right < left | ||
| my_sentence[left], my_sentence[right] = my_sentence[right], my_sentence[left] | ||
| left += 1 | ||
| right -= 1 | ||
| end |
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Question, since you reverse things twice in this method could you dry things up a bit by making a helper method?
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| regex = /([^ ]+)/ | ||
| reversed_words = my_sentence.scan(regex).flatten | ||
| # positions of regex matches with enum_for code originally from "Sean Hill"'s answer at | ||
| # https://stackoverflow.com/questions/5241653/ruby-regex-match-and-get-positions-of | ||
| # gets positions of starts of each matched reversed word | ||
| reversed_word_start_indices = my_sentence.enum_for(:scan, regex).map { Regexp.last_match.begin(0) } | ||
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| # for each start of word: | ||
| reversed_word_start_indices.each_with_index do |word_start_index, match_index| |
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n^2) loop within loop, worst case same length | ||
| # Space complexity: O(1) - only creates i and j variables each time no matter input length |
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Technically O(n) since you are doing .split
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Sorting & Reverse Sentence