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| Original file line number | Diff line number | Diff line change |
|---|---|---|
| @@ -1,13 +1,68 @@ | ||
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| require 'pry' | ||
| # Time Complexity: O(n) - as array size grows, the evaluation grows at the same rate | ||
| # Space Complexity: O(1) - no new data structures are being made | ||
| def remove_duplicates(list) | ||
| raise NotImplementedError, "Not implemented yet" | ||
| if list.length == 1 | ||
| return list | ||
| end | ||
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| i = 0 | ||
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| list.each do |number| | ||
| if number[i+1] == number[i] | ||
| list.delete_at(i) | ||
| else | ||
| i += 1 | ||
| end | ||
| end | ||
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| return list | ||
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| end | ||
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| # Time Complexity: ? | ||
| # Space Complexity: ? | ||
| # Time Complexity: O(n) - Although there are several loops, they are both proportional to the array being tested | ||
| # Space Complexity: O(1) - no new data structures are being made | ||
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| # strings being passed in is an array of strings | ||
| def longest_prefix(strings) | ||
| raise NotImplementedError, "Not implemented yet" | ||
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| max_length = strings[0].length | ||
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| return_statement = nil | ||
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| strings.each do |string| | ||
| if string.length > 0 && string.length < max_length | ||
| max_length = string.length | ||
| elsif string.length == 0 | ||
| return_statement = "" | ||
| end | ||
| end | ||
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| if return_statement != nil | ||
| return return_statement | ||
| end | ||
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| i = 0 # I am the word you're on | ||
| j = 0 # I am the character you're on | ||
| k = 0 # I am the last characer that matched | ||
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| # binding.pry | ||
| until j == max_length || i + 1 == strings.length do | ||
|
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. The order you are looking at things won't work for everything, consider:
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| if strings[i][j] == strings[i + 1][j] | ||
| i += 1 | ||
| j += 1 | ||
| k = j | ||
| else | ||
| k -= 1 | ||
| break | ||
| end | ||
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| end | ||
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| if k < 0 | ||
| return "" | ||
| else | ||
| return_statement = strings[0][0...k] | ||
| return return_statement | ||
| end | ||
| end | ||
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Since
delete_atshifts all subsequent letters over one value to the left, and has a time complexity of O(n), your algorithm is O(n)