Yasmin- Leaves #34
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Yasmin- Leaves #34
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CheezItMan
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Nov 14, 2019
| # Time complexity: ? | ||
| # 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?
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: | ||
| # Space complexity: | ||
| def reverse_inplace(s) |
| # Space complexity: ? | ||
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| # Time complexity: o(n) | ||
| # Space complexity: o(n) | ||
| def bunny(n) |
| # Space complexity: ? | ||
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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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