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Cedar - Lux #32

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640851a
Implement newman-conway
k-nox May 13, 2022
68ccfb7
implement max_subarray
k-nox Jul 8, 2022
1f48bf5
add time and space complexity for newman_conway
k-nox Jul 8, 2022
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13 changes: 10 additions & 3 deletions lib/max_subarray.py
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Original file line number Diff line number Diff line change
Expand Up @@ -2,11 +2,18 @@
def max_sub_array(nums):
""" Returns the max subarray of the given list of numbers.
Returns 0 if nums is None or an empty list.
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(n) where n is len(nums)
Space Complexity: O(1)
Comment on lines +5 to +6
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@anselrognlie anselrognlie Jul 13, 2022

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✨ 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.

"""
if nums == None:
return 0
if len(nums) == 0:
return 0
pass

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])
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@anselrognlie anselrognlie Jul 13, 2022

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✨ This is a really nice way to represent this calculation, which captures the underlying invariant that makes Kadane's algorithm work.

max_so_far = max(max_so_far, curr_max)
return max_so_far

23 changes: 17 additions & 6 deletions lib/newman_conway.py
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@@ -1,10 +1,21 @@


# Time complexity: ?
# Space Complexity: ?
def newman_conway(num):
""" Returns a list of the Newman Conway numbers for the given value.
Time Complexity: ?
Space Complexity: ?
"""Returns a list of the Newman Conway numbers for the given value.
Time Complexity: O(n) where n is num
Space Complexity: O(n) where n is num
Comment on lines +5 to +6
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@anselrognlie anselrognlie Jul 13, 2022

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✨ 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).

"""
pass
if num <= 0:
raise ValueError

if num == 1:
return "1"

nums = [0, 1, 1]
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@anselrognlie anselrognlie Jul 13, 2022

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✨ Nice use of a buffer slot to account for the 1-based calculation.


i = 3
while i <= num:
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@anselrognlie anselrognlie Jul 13, 2022

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👀 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:]])
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@anselrognlie anselrognlie Jul 13, 2022

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✨ 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))