-
Notifications
You must be signed in to change notification settings - Fork 51
Jiajia(Grace) Wang #39
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
anselrognlie
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.
✨ Your implementations look good, Jiajia! I left some comments on your implementation below.
Because of the importance of thinking about complexity for this project, I've evaluated this as a yellow due to the missing complexities for the Largest Sum Contiguous Subarray problem (wave 02). A yellow is a passing score so resubmission is not required, but you are free to resubmit with that time and space complexity filled out for a green score.
🟡
| Time Complexity: o(n) | ||
| Space 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.
✨ Great! By carefully building up the calculations and storing them for later use, we only need to perform O(n) calculations. The storage to keep those calculations is related to n (as is the converted string) giving space complexity of O(n) as well (ignoring a little bit of fiddliness related to the length of larger numbers being longer strings).
| if num == 0: | ||
| raise ValueError("n must be > 0") |
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.
We should raise this error for any value below the valid starting point of the sequence:
if num <= 0:
raise ValueError("n must be > 0")| Space Complexity: o(n) | ||
| """ | ||
| pass | ||
| nc_list = [0] * (num + 1) |
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.
✨ Nice use of a buffer slot to account for the 1-based calculation.
|
|
||
| nc_list[i] = nc_list[nc_list[i-1]] + nc_list[i - nc_list[i-1]] | ||
|
|
||
| return " ".join(str(v) for v in nc_list[1:]) |
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.
✨ Nice use of a generator to convert the numeric results to strings. This is a generator rather than a list comprehension because it lacks the [] around the comprehension expression. A generator produces a sequence of values (here, the stringified sequence values) and can be used anywhere an iterable value is needed.
Another approach would be to make uses of the map function
return " ".join(map(str, nc_list[1:]))| largest_sum = [0] * len(nums) | ||
| largest_sum[0] = nums[0] | ||
|
|
||
| for i in range(1, len(nums)): |
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.
✨ Nice approach that really shows why Kadane's algorithm is really a form of dynamic programming.
👀 What's the time and space complexity of this approach? Could we make an adjustment to who we're carrying along the previous results to improve the space complexity a bit more?
| @@ -1,4 +1,4 @@ | |||
|
|
|||
| import math | |||
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 used
No description provided.