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@kelsk kelsk commented Nov 12, 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: also O(n), where a new method is stored n times in the call stack
def factorial(n)

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if s.length > 1
first_char = s[0]
last_char = s[-1]
s = reverse(s.delete_prefix(s[0]).delete_suffix(s[-1]))

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delete_prefix & delete_suffix create new arrays and copy all the individual elements over and so this line is O(n) by itself.

# Space complexity: ?
# Time complexity: The method is called n/2 times, and removing the constant we get O(n)
# Space complexity: Since the method creates a new array every time it is called and stores a method in the call stack, it takes up 2n memory, but removing the constant makes the 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.

raise NotImplementedError, "Method not implemented"
# Time complexity: O(n), similar to the reverse(s) method
# Space complexity: O(n), also similar to the reverse(s) method
def reverse_inplace(s, index = 1)

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raise NotImplementedError, "Method not implemented"
# Time complexity: O(n)
# Space complexity: O(n), even though the method is storing a new "ears" value every time it is called
def bunny(n, ears = 0)

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raise NotImplementedError, "Method not implemented"
# Time complexity: The method will be called n/2 (or n/2 + 1) times, but since we remove the constant it's O(n)
# Space complexity: O(n) (same reasoning as above)
def nested(s, i = 1)

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def search(array, value, i = 0)
if array[i] == value
return true
elsif i > array.length

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Suggested change
elsif i > array.length
elsif i >= array.length

raise NotImplementedError, "Method not implemented"
# Time complexity: O(n), similar to the nested method
# Space complexity: O(n), same
def is_palindrome(s, i = 1)

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end
# Time complexity: O(n), where n is the number of digits in 'n' or 'm', whichever is smaller.
# Space complexity: O(n), same as above.
def digit_match(n, m, i = 0, result = 0)

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2 participants