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change space complexity of linked list's __len__ from O(n) to O(1) #8183

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Apr 1, 2023
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change space complexity of linked list's __len__ from O(n) to O(1) #8183

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2 changes: 1 addition & 1 deletion data_structures/linked_list/circular_linked_list.py
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Original file line number Diff line number Diff line change
Expand Up @@ -24,7 +24,7 @@ def __iter__(self) -> Iterator[Any]:
break

def __len__(self) -> int:
return len(tuple(iter(self)))
return sum(1 for _ in self)

def __repr__(self):
return "->".join(str(item) for item in iter(self))
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2 changes: 1 addition & 1 deletion data_structures/linked_list/doubly_linked_list.py
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Original file line number Diff line number Diff line change
Expand Up @@ -51,7 +51,7 @@ def __len__(self):
>>> len(linked_list) == 5
True
"""
return len(tuple(iter(self)))
return sum(1 for _ in self)

def insert_at_head(self, data):
self.insert_at_nth(0, data)
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2 changes: 1 addition & 1 deletion data_structures/linked_list/merge_two_lists.py
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Original file line number Diff line number Diff line change
Expand Up @@ -44,7 +44,7 @@ def __len__(self) -> int:
>>> len(SortedLinkedList(test_data_odd))
8
"""
return len(tuple(iter(self)))
return sum(1 for _ in self)

def __str__(self) -> str:
"""
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2 changes: 1 addition & 1 deletion data_structures/linked_list/singly_linked_list.py
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Original file line number Diff line number Diff line change
Expand Up @@ -72,7 +72,7 @@ def __len__(self) -> int:
>>> len(linked_list)
0
"""
return len(tuple(iter(self)))
return sum(1 for _ in self)
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The sum will call iter(self) inside, so this code is still O(n).

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For the classical linked list, the best length is O(n).

PS. collections.deque is not a real linked list, it's a wrapper around it, so it basically counts how many append/delete methods were called. https://github.com/python/cpython/blob/main/Modules/_collectionsmodule.c#L196

https://wiki.python.org/moin/TimeComplexity

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@amirsoroush amirsoroush Mar 17, 2023 •
edited
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Are you talking about "Time" complexity? This line would fix the "Space" complexity as it only fetches one item at a time and adds 1 to the final result as opposed to tuple() which loads all the items into the memory.

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I missed that, you definitely right.

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@cclauss cclauss Apr 1, 2023 •
edited
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Fun exercise... Go to https://pyodide.org/en/latest/console.html to get a current Python repl running on WASM.

Paste in the following code and hit return.

>>> from timeit import timeit
setup="from itertools import product; from string import ascii_letters"
timeit("sum(1 for _ in product(ascii_letters, repeat=4))", number=10, setup=setup)
timeit("len(tuple(product(ascii_letters, repeat=4)))", number=10, setup=setup)

5.0610000000000355
4.121999999999957

sum() is slower than len() for 7,311,616 items.

Refresh the webpage to clear out any clutter in memory...

Paste in the following code and hit return.

>>> from timeit import timeit
setup="from itertools import product; from string import ascii_letters"
timeit("sum(1 for _ in product(ascii_letters, repeat=5))", number=1, setup=setup)
timeit("len(tuple(product(ascii_letters, repeat=5)))", number=1, setup=setup)

26.686000000000035
Traceback (most recent call last):
  ...
MemoryError

sum() delivers an answer for 380,204,032 items while len() raises a MemoryError.

These numbers are for long iterators but still good to know.


def __repr__(self) -> str:
"""
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