Earth - Ringo #22
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Earth - Ringo #22
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CheezItMan
reviewed
Apr 7, 2021
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| # Time Complexity: O(1), will take same amount of time regardless of the length of the list | ||
| # Does not have to perform any additional operations to move the existing nodes | ||
| # Space Complexity: O(1), needs space for one new node | ||
| # Does not need any additional space for existing nodes as they are unaffected | ||
| # Would be O(n) where n is the size of the value being saved to the new node, | ||
| # but if all nodes are integers and therefore comparable sizes, n is constant | ||
| def add_first(value) |
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| # Time Complexity: O(n), in the worst case scenario needs to check every single | ||
| # node in a linked list of n length | ||
| # Space Complexity: O(n), just needs to save one node to a variable | ||
| def search(value) |
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👍 , but the space complexity for the iterative solution is O(1)
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| # Time Complexity: O(n). Will need to check every node in linked list of n length | ||
| # Space Complexity: O(1). Will need two running variables regardless of length of list | ||
| def find_max |
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| # Time Complexity: O(n). Will need to check every node in linked list of n length | ||
| # Space Complexity: O(1). Will need two running variables regardless of length of list | ||
| def find_min |
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| # Time Complexity: O(1), checking the head node takes one operation | ||
| # Space Complexity: O(1) assuming all nodes have comparable data sizes, same size return value regardless | ||
| def get_first |
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| # Time Complexity: O(n), n = length of linked list | ||
| # Space Complexity: O(1) | ||
| def get_last |
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| # Time Complexity: O(n), n = length of linked list | ||
| # Space Complexity: O(1) | ||
| def find_middle_value |
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👍 , good use of a fast and slow pointer
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| # Time Complexity: O(n), n = length of linked list | ||
| # Space Complexity: O(1) | ||
| def find_nth_from_end(n) |
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| # Time Complexity: O(n), n = length of linked list | ||
| # Space Complexity: O(n), n = number of nodes (length of linked list) | ||
| def has_cycle |
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That is one solution. However since you are using .include? this solution is O(n^2) in time complexity.
You can do it in O(n) with a fast and slow pointer.
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| # Time Complexity: O(n) | ||
| # Space Complexity: O(1) | ||
| def insert_ascending(value) |
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