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@brikemp brikemp commented Nov 13, 2019

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@CheezItMan CheezItMan left a comment

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Overall nice work, you hit the learning goals here. Well done. Check my comments below especially with regard to time/space complexity. Let me know if you have questions.

# Space complexity: ?
# Time complexity: O(n)
# Space complexity: O(n)
def factorial(n)

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👍

temp_char = s[0]
s[0] = s[-1]
s[-1] = temp_char
return s[0] + reverse(s[1...-1]) + s[-1]

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s[1..-1] creates a new array and copies all the individual elements over and so is O(n) by itself.


# Time complexity: O(n/2) >> O(n)
# Space complexity: O(n)
def reverse(s)

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👍
This works, but because you create a new array with each recursive call this is O(n2) for both time/space complexity.

# Space complexity: ?
# Time complexity: O(n/2) >> O(n)
# Space complexity: O(n)
def reverse_inplace(s)

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  1. This is not in place.
  2. This has similar time/space complexity to the above reverse... as in the exact same.

# Space complexity: ?
# Time complexity: O(n)
# Space complexity: O(n)
def bunny(n)

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👍

# Space complexity: ?
# Time complexity: O(n)
# Space complexity: O(n)
def nested(s)

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👍 This works, but you have similar time/space issues with the above methods.

# Space complexity: ?
# Time complexity: O(n) - worst case
# Space complexity: O(n) - worst case
def search(array, value)

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👍 This works, but you have similar time/space issues with the above methods.

# Space complexity: ?
# Time complexity: O(n/2) => O(n)
# Space complexity: O(n/2) => O(n)
def is_palindrome(s)

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👍 This works, but you have similar time/space issues with the above methods.

# Space complexity: ?
# Time complexity: O(n) n is the length of the shortest array
# Space complexity: O(n)
def digit_match(n, m)

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👍 Unusual approach, bu tit works.

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2 participants