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7 changes: 7 additions & 0 deletions .vscode/settings.json
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
}
22 changes: 17 additions & 5 deletions lib/max_subarray.py
Original file line number Diff line number Diff line change
@@ -1,12 +1,24 @@

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: ?
"""
Returns 0 if nums is None or an empty list.
Time Complexity: O(n)
Space Complexity: O(1)
Comment on lines +6 to +7

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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
if len(nums) == 1:
return nums[0]
max_sum = nums[0]
current_sum = nums[0]
for i in range(1, len(nums)):
if current_sum + nums[i] > nums[i]:
current_sum += nums[i]
else:
current_sum = nums[i]
Comment on lines +18 to +21

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

Consider using max to write it as

        current_sum = max(current_sum + nums[i], nums[i])

if current_sum > max_sum:
max_sum = current_sum
return max_sum
16 changes: 13 additions & 3 deletions lib/newman_conway.py
Original file line number Diff line number Diff line change
@@ -1,10 +1,20 @@


# Time complexity: ?
# Space Complexity: ?


def newman_conway(num):
""" Returns a list of the Newman Conway numbers for the given value.
Time Complexity: ?
Space Complexity: ?
Comment on lines 7 to 8

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👀 Time and space complexity?

"""
pass
nc_nums = [1, 1]
if num < 1:
raise ValueError("Num must be greater than 0.")
for i in range(2, num + 1):

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Since you're always truncating the returned list using num as the end, we only need to calculate up to the same limit (num) here.

nc_nums.append(nc_nums[nc_nums[i - 1] - 1] + nc_nums[i - nc_nums[i - 1]])
string_list = []
for n in nc_nums:
string_list.append(str(n))
Comment on lines +15 to +17

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Consider using a comprehension or map to apply the str transformation to each numeric value.

    string_list = [str(n) for n in nc_nums]

or

    string_list = map(str, nc_nums)

return " ".join(string_list[0:num])

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The need to slice the array here comes from needing to handle the base case for num being 1 (you initialize your calculation array to start with [1, 1], or num = 2). If you added an earlier check for the case of num being 1, you could omit the slicing here to be

    return " ".join(string_list)


# pass