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15 changes: 11 additions & 4 deletions lib/max_subarray.py
Original file line number Diff line number Diff line change
@@ -1,12 +1,19 @@

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)?
Space Complexity: constant?
Comment on lines +4 to +5

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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

current_max_array = nums[0]
max_sub_array = nums[0]

for i in range(1, len(nums)):
max_sub_array = max(max_sub_array + nums[i], nums[i])

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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.

current_max_array = max(current_max_array, max_sub_array)

return current_max_array
22 changes: 15 additions & 7 deletions lib/newman_conway.py
Original file line number Diff line number Diff line change
@@ -1,10 +1,18 @@


# Time complexity: ?
# Space Complexity: ?
def newman_conway(num):
""" Returns a list of the Newman Conway numbers for the given value.
Time Complexity: ?
Space Complexity: ?
Time Complexity: O(n)?
Space Complexity: O(n)?
Comment on lines +3 to +4

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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 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 < 1:
raise ValueError("Invalid newman conway number")
if num == 1:
return "1"

n = [0, 1, 1]

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


count = 3
while count <= num:

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👀 Prefer a for loop, since we know exactly how many times this will loop.

n.append(n[n[count - 1]] + n[count - n[count - 1]])
count += 1

return ' '.join([str(item) for item in n[1:]])

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✨ Nice use of a list comprehension to convert the numeric values to strings.

Another approach would be to use the map function:

    return ' '.join(map(str, n[1:]))