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14 changes: 14 additions & 0 deletions .vscode/settings.json
Original file line number Diff line number Diff line change
@@ -0,0 +1,14 @@
{
"python.testing.unittestArgs": [
"-v",
"-s",
"./tests",
"-p",
"*test.py"
],
"python.testing.pytestEnabled": true,
"python.testing.unittestEnabled": false,
"python.testing.pytestArgs": [
"tests"
]
}
15 changes: 12 additions & 3 deletions lib/max_subarray.py
Original file line number Diff line number Diff line change
Expand Up @@ -2,11 +2,20 @@
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: On
Space Complexity: ON
Comment on lines +5 to +6

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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).

For the space complexity, even though we are working with an input list of size n, that's not part of our space usage (it was passed in to us). The only space we use is a few numbers to calculate the running maximum (and iterate), so like the dynamic programming approach to calculating the fibonacci sequence, we are able to maintain a sliding window of recent values to complete our calculation. This lets us implement this with a constant O(1) amount of storage.

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

max_sub_array = nums[0]
max_sub = nums[0]

for i in range(1,len(nums)):

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Another way we could skip over the first value is to slice the array, or use islice from itertools to avoid making a copy (be sure to add from itertools import islice):

    for num in islice(nums, 1, None):

Then in the loop, replace nums[i] with num.


max_sub = max(max_sub + nums[i], nums[i])
max_sub_array = max(max_sub_array, max_sub)

return max_sub_array
30 changes: 26 additions & 4 deletions lib/newman_conway.py
Original file line number Diff line number Diff line change
@@ -1,10 +1,32 @@


import array

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👀 Unused import

# 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 +7 to +8

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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
# to store values
if num < 1:
raise ValueError ("Invalid Num")
if num == 1:
return "1"

f = [0, 1, 1]

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

output = "1"

count = 3
# To store values of sequence in array

while count <= num:

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

f.append(f[f[count - 1]]+ f[count - f [ count -1]])
count += 1
number = [str(item) for item in f]

return " ".join(number[1:])
Comment on lines +25 to +27

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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, number[1:]))

# for i in range(3, num + 1):
# r = f[f[i-1]]+f[i-f[i-1]]
# f.append(r);

# return r