Yasmin- Leaves #34
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Yasmin- Leaves #34
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CheezItMan
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Nov 14, 2019
CheezItMan
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CheezItMan
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As you pointed out you didn't get them all done, but most of what you have is pretty good. We'll go over the solutions in class. Let me know what questions you have.
| # Space complexity: ? | ||
| # Time complexity: o(n) | ||
| # Space complexity: o(n) | ||
| def factorial(n) |
| if s.length <= 1 | ||
| return s | ||
| else | ||
| return s[-1] + reverse(s[0..-2]) |
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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.
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| # Time complexity: o(n^2) | ||
| # Space complexity: o(n^2) | ||
| def reverse(s) |
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Correct on the time/space complexities. Can you think of a way to do better?
| # Space complexity: ? | ||
| # Time complexity: | ||
| # Space complexity: | ||
| def reverse_inplace(s) |
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| # Time complexity: o(n) | ||
| # Space complexity: o(n) | ||
| def bunny(n) |
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| # Time complexity: o(n) | ||
| # Space complexity: o(n^2) | ||
| def nested(s) |
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👍 Time complexity is also O(n^2) because of s[1...-1]
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| def search(array, value) |
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👍 Big-O is O(n^2) due to the repeated creation of new arrays.
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