Skip to content

Conversation

@emilyvomacka
Copy link

No description provided.

Copy link

@CheezItMan CheezItMan left a comment

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Nice work, both methods work. See my notes on time complexity. Well done.

(1...length).each do |k|
if list[k] == list[k-1]
i = k
until i > length || list[i-1] == nil do

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

I would say until i >= length || list[i-1] == nil do


# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n^2)

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

Correct!


# Time Complexity: ?
# Space Complexity: ?
# Time Complexity: O(n), where n is total length of all strings combined

Choose a reason for hiding this comment

The reason will be displayed to describe this comment to others. Learn more.

I would say O(n * m) where n is the number of strings and m is the number of letters in the average string. Unless the strings are small in which case you could say O(n)

Sign up for free to join this conversation on GitHub. Already have an account? Sign in to comment

Labels

None yet

Projects

None yet

Development

Successfully merging this pull request may close these issues.

2 participants