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Leaves - Alice #22
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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,13 +1,36 @@ | ||
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n), linear, because the amount of times the while loop iterates is directly proportionate to the length of the array. | ||
| # Space Complexity: O(1), constant, because there is no additional data being stored in addition to the array that is passed through, the reference itself is being changed. | ||
| def remove_duplicates(list) | ||
| raise NotImplementedError, "Not implemented yet" | ||
| i = 0 | ||
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| while i < list.length | ||
| if list[i] == list[i+1] | ||
| list.delete_at(i) | ||
| else | ||
| i+=1 | ||
| end | ||
| end | ||
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| return list | ||
| end | ||
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n*m), quadratic, because there are two loops present: the while and the each. Each of those loops are based on different data sizes, and are proportionate to the size of the data. | ||
| # Space Complexity: O(n), linear, because longest_prefix is a variable that may increase if the data size increases. | ||
| def longest_prefix(strings) | ||
|
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. Correct! |
||
| raise NotImplementedError, "Not implemented yet" | ||
| longest_prefix = "" | ||
| base_word = strings.min_by {|word| word.length} | ||
| i = 0 | ||
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| while i < base_word.length | ||
| letter_to_match = base_word[i] | ||
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| strings.each do |word| | ||
| if word[i] != letter_to_match | ||
| return longest_prefix | ||
| end | ||
| end | ||
| longest_prefix += base_word[i] | ||
| i+=1 | ||
| end | ||
| end | ||
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Since you are using
delete_atwhich has a runtime of O(n), you actually have O(n2) runtime complexity.Remember
delete_atshifts each of the subsequent elements over one index.