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CheezItMan
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CheezItMan
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Nice work, you got the methods you wrote working well. However you need to see my notes on time complexity. Remember each recursive call puts things on the system stack.
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| def fib(n) |
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This works, but see my video on time and space complexity.
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| end | ||
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| # Time complexity: O(n), where n isnthe length of the input |
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This is actually O(n^2) because s[0...-1] creates a new string and you do it n times.
| if s.nil? || s.length == 0 || s.length == 1 | ||
| return s | ||
| else | ||
| return s[-1] + reverse(s[0...-1]) |
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This is not in place since you create a new string.
Think about making a helper which takes the string, a left index and right index. The base case is when left_index >= right_index Each recursive call you swap the characters at left_index or right_index and then make a recursive call with the string, and left_index + 1 and right_index -1.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(1) because additional space in memory is not being consumed |
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O(n) since you use the system stack for the recursion.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) where n is the size of the input | ||
| # Space complexity: O(1) because additional space in memory is not being consumed |
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O(n^2) for both time and space since you use the system stack and create a new string with each iteration.
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