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CheezItMan
reviewed
Nov 14, 2019
| if s.length <= 1 | ||
| return s | ||
| else | ||
| return s[-1] + reverse(s[1...-1]) + s[0] |
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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.
| # Time complexity: ? | ||
| # 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.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(???) | ||
| # Space complexity: O(???) | ||
| def reverse_inplace(s) |
lib/recursive-methods.rb
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| # Time complexity: O(n) | ||
| # Space complexity: O(1) |
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- You are doing recursion so the system stack applies to the space complexity (at least O(n))
- Since you are doing
array.shiftthis is by itself an O(n) time operation. Combined with the recursion it's O(n2)
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(1) | ||
| def is_palindrome(s) |
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👍 This works, but you have similar time/space issues with the above methods.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n), where n is the longer number | ||
| # Space complexity: O(n) | ||
| def digit_match(n, m) |
lib/recursive-methods.rb
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| # Added Fun | ||
| # Time complexity: O(n*2^n), almost I think. |
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O(2n), actually
Space complexity O(n)
| # It'll go n-1 levels deep, with up to 2^n leaves, so n*2^n... | ||
| # Space complexity: O(n*2^n)...? not sure | ||
| def fib(n) |
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Yeah... IDK if I'm allowed to solve reverse_inplace the way I did...