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Original file line number | Diff line number | Diff line change |
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# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(n) | ||
# Space Complexity: O(1) | ||
def remove_duplicates(list) | ||
raise NotImplementedError, "Not implemented yet" | ||
index = 0 | ||
prev_num = 0 | ||
list.length.times do | ||
item = list[index] | ||
if index == 0 | ||
prev_num = item | ||
index += 1 | ||
else | ||
if item == prev_num | ||
list.delete_at(index) | ||
else | ||
index += 1 | ||
prev_num = item | ||
end | ||
end | ||
end | ||
return list | ||
end | ||
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# Time Complexity: ? | ||
# Space Complexity: ? | ||
# Time Complexity: O(a * b) where a is the number of letters in the first word, and b is the number of words in the array | ||
# Space Complexity: O(1) | ||
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. This actually doesn't work. It won't work for this. |
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raise NotImplementedError, "Not implemented yet" | ||
longest_prefix = "" | ||
num_letters = strings[0].length | ||
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num_letters.times do |i| | ||
status = true | ||
check_letter = strings[0][i] | ||
strings.each do |string| | ||
if string[i] != check_letter | ||
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. If you find a letter that won't work you can stop here because any subsequent other letters can't be in the prefix. |
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status = false | ||
end | ||
end | ||
if status == true | ||
longest_prefix += check_letter | ||
end | ||
end | ||
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return longest_prefix | ||
end | ||
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delete_at
actually shifts all the subsequent elements over 1 index. So it actually has a runtime of O(n). Since it's inside a loop running n times... Your method has a time complexity of O(n2)