Merge aa5a8858ea into fcf82a1eda

* Implemented Exponential Search with binary search for improved performance on large sorted arrays.

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

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* Added type hints and doctests for binary_search and exponential_search functions. Improved code documentation and ensured testability.

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

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* Update and rename Exponential_Search.py to exponential_search.py

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---------

Co-authored-by: pre-commit-ci[bot] <66853113+pre-commit-ci[bot]@users.noreply.github.com>

CONTRIBUTING.md

@@ -96,7 +96,7 @@ We want your work to be readable by others; therefore, we encourage you to note
 
   ```bash
   python3 -m pip install ruff  # only required the first time
-  ruff .
+  ruff check
   ```
 
 - Original code submission require docstrings or comments to describe your work.

data_structures/stacks/lexicographical_numbers.py

@@ -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))}")

dynamic_programming/longest_common_subsequence.py

@@ -28,6 +28,24 @@ def longest_common_subsequence(x: str, y: str):
     (2, 'ph')
     >>> longest_common_subsequence("computer", "food")
     (1, 'o')
+    >>> longest_common_subsequence("", "abc")  # One string is empty
+    (0, '')
+    >>> longest_common_subsequence("abc", "")  # Other string is empty
+    (0, '')
+    >>> longest_common_subsequence("", "")  # Both strings are empty
+    (0, '')
+    >>> longest_common_subsequence("abc", "def")  # No common subsequence
+    (0, '')
+    >>> longest_common_subsequence("abc", "abc")  # Identical strings
+    (3, 'abc')
+    >>> longest_common_subsequence("a", "a")  # Single character match
+    (1, 'a')
+    >>> longest_common_subsequence("a", "b")  # Single character no match
+    (0, '')
+    >>> longest_common_subsequence("abcdef", "ace")  # Interleaved subsequence
+    (3, 'ace')
+    >>> longest_common_subsequence("ABCD", "ACBD")  # No repeated characters
+    (3, 'ABD')
     """
     # find the length of strings
 

maths/trapezoidal_rule.py

@@ -1,17 +1,35 @@
 """
 Numerical integration or quadrature for a smooth function f with known values at x_i
 
-This method is the classical approach of suming 'Equally Spaced Abscissas'
+This method is the classical approach of summing 'Equally Spaced Abscissas'
 
-method 1:
-"extended trapezoidal rule"
+Method 1:
+"Extended Trapezoidal Rule"
 
 """
 
 
 def method_1(boundary, steps):
-    # "extended trapezoidal rule"
-    # int(f) = dx/2 * (f1 + 2f2 + ... + fn)
+    """
+    This function implements the extended trapezoidal rule for numerical integration.
+    The function f(x) is provided below.
+
+    :param boundary: List containing the lower and upper bounds of integration [a, b]
+    :param steps: The number of steps (intervals) used in the approximation
+    :return: The numerical approximation of the integral
+
+    >>> abs(method_1([0, 1], 10) - 0.33333) < 0.01
+    True
+
+    >>> abs(method_1([0, 1], 100) - 0.33333) < 0.01
+    True
+
+    >>> abs(method_1([0, 2], 1000) - 2.66667) < 0.01
+    True
+
+    >>> abs(method_1([1, 2], 1000) - 2.33333) < 0.01
+    True
+    """
     h = (boundary[1] - boundary[0]) / steps
     a = boundary[0]
     b = boundary[1]
@@ -19,29 +37,67 @@ def method_1(boundary, steps):
     y = 0.0
     y += (h / 2.0) * f(a)
     for i in x_i:
-        # print(i)
         y += h * f(i)
     y += (h / 2.0) * f(b)
     return y
 
 
 def make_points(a, b, h):
+    """
+    Generates the points between a and b with spacing h for trapezoidal integration.
+
+    :param a: The lower bound of integration
+    :param b: The upper bound of integration
+    :param h: The step size
+    :yield: The next x-value in the range (a, b)
+
+    >>> list(make_points(0, 1, 0.1))
+    [0.1, 0.2, 0.30000000000000004, \
+0.4, 0.5, 0.6, 0.7, \
+0.7999999999999999, \
+0.8999999999999999]
+    """
     x = a + h
     while x < (b - h):
         yield x
         x = x + h
 
 
-def f(x):  # enter your function here
-    y = (x - 0) * (x - 0)
+def f(x):
+    """
+    This is the function to integrate, f(x) = (x - 0)^2 = x^2.
+
+    :param x: The input value
+    :return: The value of f(x)
+
+    >>> f(0)
+    0.0
+
+    >>> f(1)
+    1.0
+
+    >>> f(0.5)
+    0.25
+    """
+    y = float((x - 0) * (x - 0))
     return y
 
 
 def main():
-    a = 0.0  # Lower bound of integration
-    b = 1.0  # Upper bound of integration
-    steps = 10.0  # define number of steps or resolution
-    boundary = [a, b]  # define boundary of integration
+    """
+    Main function to test the trapezoidal rule.
+    :a: Lower bound of integration
+    :b: Upper bound of integration
+    :steps: define number of steps or resolution
+    :boundary: define boundary of integration
+
+    >>> main()
+    y = 0.3349999999999999
+    """
+    a = 0.0
+    b = 1.0
+    steps = 10.0
+    boundary = [a, b]
     y = method_1(boundary, steps)
     print(f"y = {y}")
 

searches/exponential_search.py

@@ -0,0 +1,113 @@
+#!/usr/bin/env python3
+
+"""
+Pure Python implementation of exponential search algorithm
+
+For more information, see the Wikipedia page:
+https://en.wikipedia.org/wiki/Exponential_search
+
+For doctests run the following command:
+python3 -m doctest -v exponential_search.py
+
+For manual testing run:
+python3 exponential_search.py
+"""
+
+from __future__ import annotations
+
+
+def binary_search_by_recursion(
+    sorted_collection: list[int], item: int, left: int = 0, right: int = -1
+) -> int:
+    """Pure implementation of binary search algorithm in Python using recursion
+
+    Be careful: the collection must be ascending sorted otherwise, the result will be
+    unpredictable.
+
+    :param sorted_collection: some ascending sorted collection with comparable items
+    :param item: item value to search
+    :param left: starting index for the search
+    :param right: ending index for the search
+    :return: index of the found item or -1 if the item is not found
+
+    Examples:
+    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 0, 0, 4)
+    0
+    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 15, 0, 4)
+    4
+    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 5, 0, 4)
+    1
+    >>> binary_search_by_recursion([0, 5, 7, 10, 15], 6, 0, 4)
+    -1
+    """
+    if right < 0:
+        right = len(sorted_collection) - 1
+    if list(sorted_collection) != sorted(sorted_collection):
+        raise ValueError("sorted_collection must be sorted in ascending order")
+    if right < left:
+        return -1
+
+    midpoint = left + (right - left) // 2
+
+    if sorted_collection[midpoint] == item:
+        return midpoint
+    elif sorted_collection[midpoint] > item:
+        return binary_search_by_recursion(sorted_collection, item, left, midpoint - 1)
+    else:
+        return binary_search_by_recursion(sorted_collection, item, midpoint + 1, right)
+
+
+def exponential_search(sorted_collection: list[int], item: int) -> int:
+    """
+    Pure implementation of an exponential search algorithm in Python.
+    For more information, refer to:
+    https://en.wikipedia.org/wiki/Exponential_search
+
+    Be careful: the collection must be ascending sorted, otherwise the result will be
+    unpredictable.
+
+    :param sorted_collection: some ascending sorted collection with comparable items
+    :param item: item value to search
+    :return: index of the found item or -1 if the item is not found
+
+    The time complexity of this algorithm is O(log i) where i is the index of the item.
+
+    Examples:
+    >>> exponential_search([0, 5, 7, 10, 15], 0)
+    0
+    >>> exponential_search([0, 5, 7, 10, 15], 15)
+    4
+    >>> exponential_search([0, 5, 7, 10, 15], 5)
+    1
+    >>> exponential_search([0, 5, 7, 10, 15], 6)
+    -1
+    """
+    if list(sorted_collection) != sorted(sorted_collection):
+        raise ValueError("sorted_collection must be sorted in ascending order")
+
+    if sorted_collection[0] == item:
+        return 0
+
+    bound = 1
+    while bound < len(sorted_collection) and sorted_collection[bound] < item:
+        bound *= 2
+
+    left = bound // 2
+    right = min(bound, len(sorted_collection) - 1)
+    return binary_search_by_recursion(sorted_collection, item, left, right)
+
+
+if __name__ == "__main__":
+    import doctest
+
+    doctest.testmod()
+
+    # Manual testing
+    user_input = input("Enter numbers separated by commas: ").strip()
+    collection = sorted(int(item) for item in user_input.split(","))
+    target = int(input("Enter a number to search for: "))
+    result = exponential_search(sorted_collection=collection, item=target)
+    if result == -1:
+        print(f"{target} was not found in {collection}.")
+    else:
+        print(f"{target} was found at index {result} in {collection}.")