Ports - Kasey #14
Ports - Kasey #14
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CheezItMan
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CheezItMan
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Nice work, I left some notes, but you did a good job writing the methods. You do need to know that creating a string does add some time and space complexity to solutions.
Let me know if you have questions.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: o(n) where n is the length of the string |
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Since s[0...s.length-1] creates a new string, this algorithm is actually O(n^2)
| def reverse_inplace(s) | ||
| raise NotImplementedError, "Method not implemented" | ||
| return s if s == "" | ||
| return s[-1] + reverse_inplace(s[0...-1]) |
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This is creating new strings, so it isn't in place.
To do it in place you need to swap characters one at a time without creating a new string.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(1) (or O(n) cause it has that count to create and return?) |
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Since you have a system stack, it's O(n)
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) (well, O(n/2)) |
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Due to creating a new string it's actually O(n^2)
| raise NotImplementedError, "Method not implemented" | ||
| if array.length == 0 | ||
| return false | ||
| elsif array.length >= 1 && array[0] == value |
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You don't need the array.length >= 1
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) |
| # Space complexity: O(1) | ||
| def is_palindrome(s) | ||
| raise NotImplementedError, "Method not implemented" | ||
| if (s.length == 3 && s[0] == s[-1]) || (s.length == 2 && s[0] == s[-1]) || s.length == 0 |
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This could probably be simplified to return true if s.length <= 1
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(1) |
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Space complexity is O(n) due to the system stack space.
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