-
Notifications
You must be signed in to change notification settings - Fork 39
Ports - Jillianne #5
New issue
Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.
By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement. We’ll occasionally send you account related emails.
Already on GitHub? Sign in to your account
base: master
Are you sure you want to change the base?
Conversation
CheezItMan
left a comment
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
Not bad, you have some effective recursive methods. They could be more efficient (see my notes on reverse in place for a hint as to how), and you were off on the Big-O, but a good 1st attempt at recursive methods.
|
|
||
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
This is actually O(n^2) due tot he fact that s[0..-1] creates a new string of length n-1
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def bunny(n) |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
👍
|
|
||
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n)? |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
O(n^2) for both space and time, see above.
| # I am not sure how to approach this!! | ||
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| def reverse_inplace(s) |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
For reverse in place think about making a helper which takes the string, a left index and right index. The base case is when left_index >= right_index Each recursive call you swap the characters at left_index or right_index and then make a recursive call with the string, and left_index + 1 and right_index -1.
|
|
||
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
O(n^2) similar to the above.
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def is_palindrome(s) |
There was a problem hiding this comment.
Choose a reason for hiding this comment
The reason will be displayed to describe this comment to others. Learn more.
O(n^2) similar to the above.
I am still unsure of time and space complexity for my solutions!