The black formatter is no longer beta (#5960)

* The black formatter is no longer beta

* pre-commit autoupdate

* pre-commit autoupdate

* Remove project_euler/problem_145 which is killing our CI tests

* updating DIRECTORY.md

Co-authored-by: github-actions <${GITHUB_ACTOR}@users.noreply.github.com>

.pre-commit-config.yaml

@@ -1,6 +1,6 @@
 repos:
   - repo: https://github.com/pre-commit/pre-commit-hooks
-    rev: v3.4.0
+    rev: v4.1.0
     hooks:
       - id: check-executables-have-shebangs
       - id: check-yaml
@@ -14,26 +14,26 @@ repos:
       - id: requirements-txt-fixer
 
   - repo: https://github.com/psf/black
-    rev: 21.4b0
+    rev: 22.1.0
     hooks:
       - id: black
 
   - repo: https://github.com/PyCQA/isort
-    rev: 5.8.0
+    rev: 5.10.1
     hooks:
       - id: isort
         args:
           - --profile=black
 
   - repo: https://github.com/asottile/pyupgrade
-    rev: v2.29.0
+    rev: v2.31.0
     hooks:
       - id: pyupgrade
         args:
           - --py39-plus
 
   - repo: https://gitlab.com/pycqa/flake8
-    rev: 3.9.1
+    rev: 3.9.2
     hooks:
       - id: flake8
         args:
@@ -42,7 +42,7 @@ repos:
           - --max-line-length=88
 
   - repo: https://github.com/pre-commit/mirrors-mypy
-    rev: v0.910
+    rev: v0.931
     hooks:
       - id: mypy
         args:
@@ -51,11 +51,11 @@ repos:
           - --non-interactive
 
   - repo: https://github.com/codespell-project/codespell
-    rev: v2.0.0
+    rev: v2.1.0
     hooks:
       - id: codespell
         args:
-          - --ignore-words-list=ans,crate,fo,followings,hist,iff,mater,secant,som,tim
+          - --ignore-words-list=ans,crate,fo,followings,hist,iff,mater,secant,som,sur,tim
           - --skip="./.*,./strings/dictionary.txt,./strings/words.txt,./project_euler/problem_022/p022_names.txt"
         exclude: |
           (?x)^(

DIRECTORY.md

@@ -856,8 +856,6 @@
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_135/sol1.py)
   * Problem 144
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_144/sol1.py)
-  * Problem 145
-    * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_145/sol1.py)
   * Problem 173
     * [Sol1](https://github.com/TheAlgorithms/Python/blob/master/project_euler/problem_173/sol1.py)
   * Problem 174

project_euler/problem_145/sol1.py

@@ -1,87 +0,0 @@
-"""
-Problem 145: https://projecteuler.net/problem=145
-
-Name: How many reversible numbers are there below one-billion?
-
-Some positive integers n have the property that the
-sum [ n + reverse(n) ] consists entirely of odd (decimal) digits.
-For instance, 36 + 63 = 99 and 409 + 904 = 1313.
-We will call such numbers reversible; so 36, 63, 409, and 904 are reversible.
-Leading zeroes are not allowed in either n or reverse(n).
-
-There are 120 reversible numbers below one-thousand.
-
-How many reversible numbers are there below one-billion (10^9)?
-
-
-Solution:
-
-Here a brute force solution is used to find and count the reversible numbers.
-
-"""
-from __future__ import annotations
-
-
-def check_if_odd(sum: int = 36) -> int:
-    """
-    Check if the last digit in the sum is even or odd. If even return 0.
-    If odd then floor division by 10 is used to remove the last number.
-    Process continues until sum becomes 0 because no more numbers.
-    >>> check_if_odd(36)
-    0
-    >>> check_if_odd(33)
-    1
-    """
-    while sum > 0:
-        if (sum % 10) % 2 == 0:
-            return 0
-        sum = sum // 10
-    return 1
-
-
-def find_reverse_number(number: int = 36) -> int:
-    """
-    Reverses the given number. Does not work with number that end in zero.
-    >>> find_reverse_number(36)
-    63
-    >>> find_reverse_number(409)
-    904
-    """
-    reverse = 0
-
-    while number > 0:
-        temp = number % 10
-        reverse = reverse * 10 + temp
-        number = number // 10
-
-    return reverse
-
-
-def solution(number: int = 1000000000) -> int:
-    """
-    Loops over the range of numbers.
-    Checks if they have ending zeros which disqualifies them from being reversible.
-    If that condition is passed it generates the reversed number.
-    Then sum up n and reverse(n).
-    Then check if all the numbers in the sum are odd. If true add to the answer.
-    >>> solution(1000000000)
-    608720
-    >>> solution(1000000)
-    18720
-    >>> solution(1000000)
-    18720
-    >>> solution(1000)
-    120
-    """
-    answer = 0
-    for x in range(1, number):
-        if x % 10 != 0:
-            reversed_number = find_reverse_number(x)
-            sum = x + reversed_number
-            answer += check_if_odd(sum)
-
-    return answer
-
-
-if __name__ == "__main__":
-    print(f"{solution() = }")