[pre-commit.ci] pre-commit autoupdate (#11322)

* [pre-commit.ci] pre-commit autoupdate

updates:
- [github.com/astral-sh/ruff-pre-commit: v0.2.2 → v0.3.2](https://github.com/astral-sh/ruff-pre-commit/compare/v0.2.2...v0.3.2)
- [github.com/pre-commit/mirrors-mypy: v1.8.0 → v1.9.0](https://github.com/pre-commit/mirrors-mypy/compare/v1.8.0...v1.9.0)

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

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

---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>
This commit is contained in:
pre-commit-ci[bot]
2024-03-13 07:52:41 +01:00
committed by GitHub
parent 5f95d6f805
commit bc8df6de31
297 changed files with 488 additions and 285 deletions

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@ -5,6 +5,7 @@ python3 -m doctest -v avl_tree.py
For testing run:
python avl_tree.py
"""
from __future__ import annotations
import math

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@ -88,6 +88,7 @@ True
>>> not t
True
"""
from __future__ import annotations
from collections.abc import Iterable, Iterator

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@ -7,6 +7,7 @@ python -m unittest binary_search_tree_recursive.py
To run an example:
python binary_search_tree_recursive.py
"""
from __future__ import annotations
import unittest

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@ -8,7 +8,6 @@ Python implementation:
frames that could be in memory is `n`
"""
from __future__ import annotations
from collections.abc import Iterator

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@ -2,6 +2,7 @@
The diameter/width of a tree is defined as the number of nodes on the longest path
between two end nodes.
"""
from __future__ import annotations
from dataclasses import dataclass

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@ -10,6 +10,7 @@ https://www.geeksforgeeks.org/flatten-a-binary-tree-into-linked-list
Author: Arunkumar A
Date: 04/09/2023
"""
from __future__ import annotations

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@ -9,6 +9,7 @@ https://bit.ly/46uB0a2
Author : Arunkumar
Date : 14th October 2023
"""
from __future__ import annotations
from collections.abc import Iterator

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@ -13,6 +13,7 @@ If n is the number of nodes in the tree then:
Runtime: O(n)
Space: O(1)
"""
from __future__ import annotations
from collections.abc import Iterator

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@ -3,6 +3,7 @@ Is a binary tree a sum tree where the value of every non-leaf node is equal to t
of the values of its left and right subtrees?
https://www.geeksforgeeks.org/check-if-a-given-binary-tree-is-sumtree
"""
from __future__ import annotations
from collections.abc import Iterator

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@ -5,6 +5,7 @@ The rule for merging is that if two nodes overlap, then put the value sum of
both nodes to the new value of the merged node. Otherwise, the NOT null node
will be used as the node of new tree.
"""
from __future__ import annotations

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@ -3,6 +3,7 @@ Given the root of a binary tree, mirror the tree, and return its root.
Leetcode problem reference: https://leetcode.com/problems/mirror-binary-tree/
"""
from __future__ import annotations
from collections.abc import Iterator

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@ -35,6 +35,7 @@ https://www.geeksforgeeks.org/segment-tree-efficient-implementation/
>>> st.query(0, 2)
[1, 2, 3]
"""
from __future__ import annotations
from collections.abc import Callable

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@ -6,6 +6,7 @@ We will use the formula: t(n) = SUMMATION(i = 1 to n)t(i-1)t(n-i)
Further details at Wikipedia: https://en.wikipedia.org/wiki/Catalan_number
"""
"""
Our Contribution:
Basically we Create the 2 function:

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@ -2,6 +2,7 @@
psf/black : true
ruff : passed
"""
from __future__ import annotations
from collections.abc import Iterator

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@ -3,6 +3,7 @@ Segment_tree creates a segment tree with a given array and function,
allowing queries to be done later in log(N) time
function takes 2 values and returns a same type value
"""
from collections.abc import Sequence
from queue import Queue

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@ -4,6 +4,7 @@ Given the root of a binary tree, check whether it is a mirror of itself
Leetcode reference: https://leetcode.com/problems/symmetric-tree/
"""
from __future__ import annotations
from dataclasses import dataclass

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@ -7,6 +7,7 @@ such as the with segment trees or fenwick trees. You can read more about them he
2. https://www.youtube.com/watch?v=4aSv9PcecDw&t=811s
3. https://www.youtube.com/watch?v=CybAgVF-MMc&t=1178s
"""
from __future__ import annotations
test_array = [2, 1, 4, 5, 6, 0, 8, 9, 1, 2, 0, 6, 4, 2, 0, 6, 5, 3, 2, 7]