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CheezItMan
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CheezItMan
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Really nice work, you hit all the learning goals here. Take a look at my comments regarding time complexity and let me know if you have questions.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) - where n is equal to the length of the input string |
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Because s[0..-2] will also create a new array, this makes the time complexity also O(n^2)
| raise NotImplementedError, "Method not implemented" | ||
| # Time complexity: O(n) - where n is the length of the string / 2 | ||
| # Space complexity: O(n) - where n is the length of the string / 2 | ||
| def reverse_inplace(s, i = 0, j = s.length - 1) |
| raise NotImplementedError, "Method not implemented" | ||
| if s == "" | ||
| return true | ||
| elsif s.length == 1 || s[0] != "(" || s[-1] != ")" |
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if s.length ==1 then it can't have a matching paren.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) - where n equal the length of the string / 2 |
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Because s[1..-2] creates a new string of length n-2, this is an O(n^2) for both space and time.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) - where n equals the length of the array |
| # Space complexity: ? | ||
| # Time complexity: O(n) - where n equals the length of the string / 2 | ||
| # Space complexity: O(n) - where n equals the length of the string / 2 + length of the string | ||
| def is_palindrome(s) |
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O(n^2) similar to the above, look at your reverse in place, could you use a similar technique?
| def digit_match(n, m) | ||
| raise NotImplementedError, "Method not implemented" | ||
| end No newline at end of file | ||
| if (n / 10 == 0) || (m / 10 == 0) |
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These ifs could be simplified a bit. Otherwise very good!
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