Fire - Anna #17
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Fire - Anna #17
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CheezItMan
reviewed
Mar 23, 2021
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| # method to find if the linked list contains a node with specified value | ||
| # returns true if found, false otherwise | ||
| # Time Complexity: O(n) | ||
| # >> Must check every single node in linked list | ||
| # Space Complexity: O(1) | ||
| # >> Just checking for membership; only ever store current_node | ||
| def search(value) |
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| # method to add a new node with the specific data value in the linked list | ||
| # insert the new node at the beginning of the linked list | ||
| # Time Complexity: O(1) | ||
| # >> Constant num of actions - create new node, | ||
| # set next to old head, set head pointer to new node | ||
| # Space Complexity: O(1) | ||
| # >> Always adding 1 new node | ||
| def add_first(value) |
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| # method to return the max value in the linked list | ||
| # returns the data value and not the node | ||
| def find_max |
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| # Time Complexity: O(n) | ||
| # >> have to traverse entire linked list (unless we know it's sorted...) | ||
| # Space Complexity: O(1) | ||
| # >> store current_node and min regardless of how long LL is | ||
| def find_min |
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| # Time Complexity: O(n) | ||
| # >> get length of LL (pass through all n nodes) | ||
| # >> go through half of those nodes | ||
| # Space Complexity: O(1) | ||
| # >> store mid_index, current_node, current_node_num | ||
| def find_middle_value |
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| mid_index = length / 2 | ||
| current_node = @head | ||
| current_node_num = 0 |
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This works, but could you do something similar to the method to find a cycle.
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Ah I see - I think India mentioned in the previous CS Fun session that you could move a slow and fast pointer and when fast reaches the end, slow will be at the middle
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| # Time Complexity: O(n) | ||
| # >> pass through all nodes to get length | ||
| # >> at worst pass through all nodes again to get 0th node from end | ||
| # >> 2n --> n | ||
| # Space Complexity: O(1) | ||
| # >> store len, index, current_node, curr_node_num | ||
| def find_nth_from_end(n) |
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| # Time Complexity: O(n) | ||
| # >> If no cycle, 'fast' and 'slow' pointers will go through LL once | ||
| # >> If cycle, 'slow' pointer will go through the list at most once | ||
| # >> b/c 'fast' cannot lap / overtake the 'slow' pointer if 'fast' | ||
| # >> moves two nodes at a time while 'slow' 1 node at a time | ||
| # >> if DID skip over --> slow --> fast | ||
| # >> previous step would mean both pointers were at the same node | ||
| # Space Complexity: O(1) | ||
| # >> fast and slow nodes | ||
| def has_cycle |
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| # Time Complexity: O(n) | ||
| # >> worst case node has to be placed at end | ||
| # Space Complexity: O(1) | ||
| # >> store new_node and current_node | ||
| def insert_ascending(value) |
Co-authored-by: Chris M <[email protected]>
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