Angele Z. #32
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Angele Z. #32
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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) |
| # Space complexity: O(n) | ||
| def reverse_inplace(s, i = 0) | ||
| return s if i == s.length / 2 | ||
| s[i], s[s.length - i - 1] = s[s.length - i - 1], s[i] |
| raise NotImplementedError, "Method not implemented" | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def reverse_inplace(s, i = 0) |
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n) | ||
| # Space complexity: O(n) | ||
| def bunny(n) |
| end | ||
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| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n^2) since using .length? |
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.length is actually O(1) because it's an instance variable (for that explicit reason).
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n^2) since using .length? | ||
| # Space complexity: O(n) | ||
| def nested(s) |
| return false if array.empty? | ||
| return true if array[0] == value | ||
| new_array = array[1..-1] | ||
| return search(array[1..-1], value) |
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array[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 search(array, value) |
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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(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.
| # Time complexity: ? | ||
| # Space complexity: ? | ||
| # Time complexity: O(n^2) since it uses .length in each loop? | ||
| # Space complexity: O(n) | ||
| def digit_match(n, m) |
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👍 This works, but you have similar time/space issues with the above methods.
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