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Sara-Branches #25
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Sara-Branches #25
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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.
# Time complexity: ? | ||
# Space complexity: ? | ||
# Time complexity: 0(n) | ||
# Space complexity: 0(n) | ||
def factorial(n) |
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👍
raise NotImplementedError, "Method not implemented" | ||
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return s if s.length < 2 | ||
return reverse(s[1..-1]) + s[0] |
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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: ? | ||
# Space complexity: ? | ||
# Time complexity: 0(n) | ||
# Space complexity: 0(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.
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return s if s.length < 2 | ||
return reverse(s[1..-1]) + s[0] | ||
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end | ||
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# Time complexity: ? | ||
# Space complexity: ? | ||
def reverse_inplace(s) |
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- This is not done in place
- It's calling the method above.
# Time complexity: ? | ||
# Space complexity: ? |
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❓
def nested_helper(s) | ||
return true if s.empty? | ||
return false if s.length.odd? | ||
return nested_helper(s) | ||
end | ||
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# Time complexity: ? | ||
# Space complexity: ? | ||
def search(array, value) |
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👍 This works, but you have similar time/space issues with the above methods.
def is_palindrome_helper(s, index) | ||
return true if s.empty? || s.length ==1 | ||
return false if s[0] != s[s.length - index] | ||
is_palindrome_helper(s[1..s.length - index - 1], 1) |
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You're both slicing the array and using index
as 1, I'm not sure why index is needed as a param at all.
raise NotImplementedError, "Method not implemented" | ||
return false if array.nil? | ||
return true if array[0] == value | ||
return search(array[1..-1],value) | ||
end | ||
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# Time complexity: ? | ||
# Space complexity: ? | ||
def is_palindrome(s) |
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👍 This works, but you have similar time/space issues with the above methods.
if n % 10 == m % 10 | ||
count += 1 | ||
end | ||
if (n / 10 == 0 || m / 10 == 0) |
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🏆 Nice catch!
return true if s.empty? || s.length ==1 | ||
return false if s[0] != s[s.length - index] | ||
is_palindrome_helper(s[1..s.length - index - 1], 1) | ||
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end | ||
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# Time complexity: ? | ||
# Space complexity: ? | ||
def digit_match(n, m) |
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👍
No description provided.