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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) |
| 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) |
| 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) |
| 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) |
| 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) |
| end No newline at end of file | ||
| # 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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