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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) |
| # Space complexity: o(n) | ||
| def reverse(s) | ||
| raise NotImplementedError, "Method not implemented" | ||
| return reverse_recursive(s, reversed_string = "", index = -1) |
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| return reverse_recursive(s, reversed_string = "", index = -1) | |
| return reverse_recursive(s, "", index = -1) |
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| def reverse_recursive(s, reversed_string, index) | ||
| return reversed_string if index < -s.length | ||
| reversed_string += s[index] |
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reversed_string += s[index] creates a new array and copies all the individual elements over and so is O(n) by itself.
| # Space complexity: ? | ||
| # Time complexity: o(n) | ||
| # 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.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| def reverse_inplace_recursive(s, last, index) |
| raise NotImplementedError, "Method not implemented" | ||
| return true if s.empty? | ||
| return false if s[0] != "(" || s[-1] != ")" | ||
| return nested(s[1...-1]) |
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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(n) | ||
| # Space complexity: o(n) | ||
| def nested(s) |
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This works, but you have similar time/space issues with the above methods.
| # Space complexity: ? | ||
| # Time complexity: o(n) | ||
| # Space complexity: o(n) | ||
| def search(array, value) |
| # Space complexity: ? | ||
| # Time complexity: o(n) | ||
| # Space complexity: o(n) | ||
| def is_palindrome(s) |
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👍 This works, but you have similar time/space issues with the above methods.
| # Space complexity: ? | ||
| # Time complexity: o(n) | ||
| # Space complexity: o(n) | ||
| def digit_match(n, m) |
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