Compare commits

...

3 Commits

Author SHA1 Message Date
NoodlesIAte
04956d8694
Merge 8db8f5b24b into 5a8655d306 2024-10-05 10:20:17 -07:00
1227haran
5a8655d306
Added new algorithm to generate numbers in lexicographical order (#11674)
* Added algorithm to generate numbers in lexicographical order

* Removed the test cases

* Updated camelcase to snakecase

* Added doctest

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Added descriptive name for n

* Reduced the number of letters

* Updated the return type

* [pre-commit.ci] auto fixes from pre-commit.com hooks

for more information, see https://pre-commit.ci

* Updated import statement

* Updated return type to Iterator[int]

* removed parentheses

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
2024-10-05 10:19:58 -07:00
NoodlesIAte
8db8f5b24b
Added complexities 2024-10-04 10:06:58 +05:30
2 changed files with 47 additions and 0 deletions

View File

@ -0,0 +1,38 @@
from collections.abc import Iterator
def lexical_order(max_number: int) -> Iterator[int]:
"""
Generate numbers in lexical order from 1 to max_number.
>>> " ".join(map(str, lexical_order(13)))
'1 10 11 12 13 2 3 4 5 6 7 8 9'
>>> list(lexical_order(1))
[1]
>>> " ".join(map(str, lexical_order(20)))
'1 10 11 12 13 14 15 16 17 18 19 2 20 3 4 5 6 7 8 9'
>>> " ".join(map(str, lexical_order(25)))
'1 10 11 12 13 14 15 16 17 18 19 2 20 21 22 23 24 25 3 4 5 6 7 8 9'
>>> list(lexical_order(12))
[1, 10, 11, 12, 2, 3, 4, 5, 6, 7, 8, 9]
"""
stack = [1]
while stack:
num = stack.pop()
if num > max_number:
continue
yield num
if (num % 10) != 9:
stack.append(num + 1)
stack.append(num * 10)
if __name__ == "__main__":
from doctest import testmod
testmod()
print(f"Numbers from 1 to 25 in lexical order: {list(lexical_order(26))}")

View File

@ -26,6 +26,15 @@ def quick_sort(collection: list) -> list:
[]
>>> quick_sort([-2, 5, 0, -45])
[-45, -2, 0, 5]
Time Complexity:
Best Case: O(n log n)
Average case: O(n log n)
Worst Case: O(n^2)
Space Complexity:
Best Case: O(log n)
Worst Case: O(n)
"""
# Base case: if the collection has 0 or 1 elements, it is already sorted
if len(collection) < 2: