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C16 - Spruce - Vange Spracklin #45

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4d87ba8
max contiguous array
Jul 22, 2022
f24335a
base cases passing
Jul 22, 2022
252beab
newman conwa 13 nad 20
Jul 22, 2022
6aa1355
time and space complexity
Jul 22, 2022
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7 changes: 7 additions & 0 deletions .vscode/settings.json
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Original file line number Diff line number Diff line change
@@ -0,0 +1,7 @@
{
"python.testing.pytestArgs": [
"tests"
],
"python.testing.unittestEnabled": false,
"python.testing.pytestEnabled": true
}
16 changes: 13 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 +2,21 @@
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), iterates through relative to the size of the list given.
Space Complexity: ? O(1), we will only ever make 2
Comment on lines +5 to +6
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@anselrognlie anselrognlie Jul 23, 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.

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

# initialize /memoize:
max_subarray_so_far = nums[0]
max_this_subarray = nums[0]

for i in range(1, len(nums)): # O(n)
max_this_subarray = max(max_this_subarray + nums[i], nums[i])
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@anselrognlie anselrognlie Jul 23, 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.

You could also refer back to the pseudocode presented in the project description for a slightly different (though equivalent) approach. Again, to me it's not obvious why Kadane's algorithm is correct, so do check out that linked article.

max_subarray_so_far = max(max_this_subarray, max_subarray_so_far)

return max_subarray_so_far
34 changes: 27 additions & 7 deletions lib/newman_conway.py
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@@ -1,10 +1,30 @@


# Time complexity: ?
# Space Complexity: ?
'''
P(1) = 1
P(2) = 1
for all n > 2
P(n) = P(P(n - 1)) + P(n - P(n - 1))
'''
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)
"""
pass
# base case guard clauses
if num < 1:
raise ValueError('newman conway number cant be zero nums long')
elif num == 1:
return '1'

memo = [0,1,1]
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@anselrognlie anselrognlie Jul 23, 2022

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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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@anselrognlie anselrognlie Jul 23, 2022

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

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

NCnums = []
for num in memo:
NCnums.append(str(num))
Comment on lines +26 to +28
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@anselrognlie anselrognlie Jul 23, 2022

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Consider using a comprehension or map to apply the str transformation to each numeric value. And avoid using identifiers that start with capital letters for things that aren't classes.

    nc_nums = [str(num) for num in memo]

or

    nc_nums = map(str, memo)

# print(NCnums[1:])
return " ".join(NCnums[1:])