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CheezItMan
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CheezItMan
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Not bad, you wrote some effective recursive methods, but missed a bit on the time & space complexities. See my inline notes and let me know if you have questions.
| def reverse(s) | ||
| raise NotImplementedError, "Method not implemented" | ||
| return s if s.length <= 1 | ||
| reversed_str = reverse(s[1..-1]) |
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s[1..-1] creates a new string of length n-2, which means your time and space complexities are going to be O(n^2)
| def reverse_inplace(s) | ||
| raise NotImplementedError, "Method not implemented" | ||
| return s if s.length < 1 | ||
| return (s[-1] + reverse_inplace(s[0...-1])) |
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This is not in place since you create a new string.
Think about making a helper which takes the string, a left index and right index. The base case is when left_index >= right_index Each recursive call you swap the characters at left_index or right_index and then make a recursive call with the string, and left_index + 1 and right_index -1.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(1) |
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Space complexity will be O(n^2) due to the system stack and a new string on each level of the stack.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
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This is also O(n^2) for both space and time complexity.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
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O(n^2) for both time and space complexity. You could do it with O(n) using a helper method and current left_index and right_index parameters.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(1) |
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Due to the system stack, the space complexity will be O(n)
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