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Roza - cedar #22

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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^2)
Space Complexity: O(1)
Comment on lines +5 to +6
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@kyra-patton kyra-patton Jun 27, 2022

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✨ This works and your time and space complexity does match your solution, but I think you're doing a little extra work here and we can make this solution even more efficient! See comments below ⬇️

"""
if nums == None:
return 0
if len(nums) == 0:
return 0
pass
max_overall = nums[0]
max_so_far = nums[0]
for index, num in enumerate(nums):
so_far_sum = 0
for each in nums[:index + 1]:
so_far_sum += each
max_so_far = max(so_far_sum, num)
Comment on lines +15 to +18
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@kyra-patton kyra-patton Jun 27, 2022

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With your current solution, you calculate what the sum would be for values from index 0 to the current index, and set 'max_so_far' to be either that calculated value or the current value you are iterating over.

We can eliminate the need for that inner loop/calculation by instead just checking if max_so_far + the next value in the list is greater than the next value in the list. If max_so_far + num is greater than num, max_so_far in the next iteration will already hold the so_far_sum value you're currently calculating with a loop.

If max_so_far + num is less than num, that means the so_far_sum shouldn't be a part of the subarray anyways, so there's no need to calculate it in this case either.

How would this change to your code affect your time and space complexity? 🤔

Suggested change
so_far_sum = 0
for each in nums[:index + 1]:
so_far_sum += each
max_so_far = max(so_far_sum, num)
max_so_far = max(max_so_far + num, num)

max_overall = max (max_overall, max_so_far)

return max_overall

26 changes: 20 additions & 6 deletions lib/newman_conway.py
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Original file line number Diff line number Diff line change
@@ -1,10 +1,24 @@


# Time complexity: ?
# Space Complexity: ?
# Time complexity: O(n)
# Space Complexity: O(n)
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)
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@kyra-patton kyra-patton Jun 27, 2022

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✨

"""
pass
if num == 0:
raise ValueError
if num == 1:
return "1"
if num == 2:
return "1 1"
initial_array = [0, 1, 1]
for i in range(3, num + 1):
r = initial_array[initial_array[i-1]]+initial_array[i-initial_array[i-1]]
initial_array.append(r)
# remove the first 0 using list slices and change elements to str
stringified_array = []
for elem in initial_array[1:]:
stringified_array.append(str(elem))
return " ".join(stringified_array)