Leaves - Tiffany #38
Leaves - Tiffany #38
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CheezItMan
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CheezItMan
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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) |
| raise NotImplementedError, "Method not implemented" | ||
| return s if s.length <= 1 | ||
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| return s[s.length-1] << reverse(s[1..-2]) << 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.
| # Space complexity: ? | ||
| # Time complexity: O(s.length) or O(n) | ||
| # Space complexity: O(s.length) or 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.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Space complexity: O(1) | ||
| def reverse_inplace(s) |
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def bunny(n) |
| # Space complexity: ? | ||
| # Time complexity: O(array.length) or O(n) | ||
| # Space complexity: O(1) | ||
| def search(array, value) |
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👍 This works, but you have similar time/space issues with the above methods due to creating new arrays.
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| return false if s[0] != s[-1] | ||
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| is_palindrome(s[1..-2]) |
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| is_palindrome(s[1..-2]) | |
| return is_palindrome(s[1..-2]) |
| # Space complexity: ? | ||
| # Time complexity: O(s.length) or O(n) | ||
| # Space complexity: O(1) | ||
| def is_palindrome(s) |
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👍 This works, but you have similar time/space issues with the above methods due to creating new arrays.
Remember your minimum space complexity would be O(n), even without creating new arrays with each recursive call. You have to consider the system call stack.
| if n < 10 || m < 10 | ||
| return 1 |
| # Space complexity: ? | ||
| # Time complexity: O(log(n)) or O(log(m)) whichever is shorter | ||
| # Space complexity: O(log(n)) or O(log(m)) whichever is shorter | ||
| def digit_match(n, m) |
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👍 Correct where n is the size of the number, you could also say O(n) where n is the least number of digits in a number.
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