Cedar - Lux #32
Cedar - Lux #32
Conversation
anselrognlie
left a comment
anselrognlie
left a comment
There was a problem hiding this comment.
✨ 💫 Looks good, Lux! I left some comments on your implementation below.
🟢
| Time Complexity: O(n) where n is num | ||
| Space Complexity: O(n) where n is num |
There was a problem hiding this comment.
✨ 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 sliced copy, and 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 == 1: | ||
| return "1" | ||
|
|
||
| nums = [0, 1, 1] |
There was a problem hiding this comment.
✨ Nice use of a buffer slot to account for the 1-based calculation.
| nums = [0, 1, 1] | ||
|
|
||
| i = 3 | ||
| while i <= num: |
There was a problem hiding this comment.
👀 Prefer a for loop, since we know exactly how many times this will loop.
| nums.append(nums[nums[i - 1]] + nums[i - nums[i - 1]]) | ||
| i += 1 | ||
|
|
||
| return " ".join([str(n) for n in nums[1:]]) |
There was a problem hiding this comment.
✨ Nice use of a list comprehension to convert the int values to strings for use in the join.
We could also use map
return " ".join(map(str, nums[1:]))We could consider using islice (from itertools) to lop of that first character without making an actual copy:
return " ".join(map(str, islice(nums, 1, None))| Time Complexity: O(n) where n is len(nums) | ||
| Space Complexity: O(1) |
There was a problem hiding this comment.
✨ Notice how better time complexity this approach achieves over a "naïve" approach of checking for the maximum achievable sum starting from every position and every length. The correctness of this approach might not be apparent, so I definitely encourage reading a bit more about it. This has a fairly good explanation, as well as a description of why this is considered a dynamic programming approach (on the face it might not "feel" like one).
Since like the fibonacci sequence, we are able to maintain a sliding window of recent values to complete our calculation, we can do it with a constant O(1) amount of storage.
| max_so_far = nums[0] | ||
| curr_max = nums[0] | ||
| for i in range(1, len(nums)): | ||
| curr_max = max(curr_max + nums[i], nums[i]) |
There was a problem hiding this comment.
✨ This is a really nice way to represent this calculation, which captures the underlying invariant that makes Kadane's algorithm work.
2 participants
No description provided.