Leaves - Dominique #43
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CheezItMan
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Oct 10, 2019
CheezItMan
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CheezItMan
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Outstanding work! You hit all the learning goals here. Very well done!
Take a look at my comments and let me know if you have questions.
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| def word_reverse(my_sentence, word_start, word_end) | ||
| word_length = word_end - word_start + 1 | ||
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| counter = 0 | ||
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| while counter != word_length/2 | ||
| temp = my_sentence[word_start] | ||
| my_sentence[word_start] = my_sentence[word_end] | ||
| my_sentence[word_end] = temp | ||
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| word_start += 1 | ||
| word_end -= 1 | ||
| counter += 1 | ||
| end | ||
| end | ||
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| until i == my_sentence.length/2 | ||
| temp = my_sentence[i] | ||
| my_sentence[i] = my_sentence[my_sentence.length - i - 1] | ||
| my_sentence[my_sentence.length - i - 1] = temp | ||
| i += 1 | ||
| end |
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You could also use the word_reverse method again to reverse the whole string.
| # Space complexity: ? | ||
| # Time complexity: O(n^2) because as the algorithm scales, each nested loop will run n number of times according to the input. | ||
| # Space complexity: O(1) because the same amount of memory will be used regardless of the input size. | ||
| def sort_by_length(my_sentence) |
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n^2) because as the algorithm scales, each nested loop will run n number of times according to the input. | ||
| # Space complexity: O(1) because the same amount of memory will be used regardless of the input size. |
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Technically since you are doing .split, you have O(n) space complexity.
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Sorting & Reverse Sentence