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@ -96,7 +96,7 @@ We want your work to be readable by others; therefore, we encourage you to note
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```bash
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python3 -m pip install ruff # only required the first time
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ruff .
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ruff check
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```
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- Original code submission require docstrings or comments to describe your work.
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38
data_structures/stacks/lexicographical_numbers.py
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38
data_structures/stacks/lexicographical_numbers.py
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@ -0,0 +1,38 @@
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from collections.abc import Iterator
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def lexical_order(max_number: int) -> Iterator[int]:
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"""
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Generate numbers in lexical order from 1 to max_number.
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>>> " ".join(map(str, lexical_order(13)))
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'1 10 11 12 13 2 3 4 5 6 7 8 9'
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>>> list(lexical_order(1))
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[1]
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>>> " ".join(map(str, lexical_order(20)))
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'1 10 11 12 13 14 15 16 17 18 19 2 20 3 4 5 6 7 8 9'
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>>> " ".join(map(str, lexical_order(25)))
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'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'
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>>> list(lexical_order(12))
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[1, 10, 11, 12, 2, 3, 4, 5, 6, 7, 8, 9]
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"""
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stack = [1]
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while stack:
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num = stack.pop()
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if num > max_number:
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continue
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yield num
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if (num % 10) != 9:
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stack.append(num + 1)
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stack.append(num * 10)
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if __name__ == "__main__":
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from doctest import testmod
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testmod()
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print(f"Numbers from 1 to 25 in lexical order: {list(lexical_order(26))}")
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@ -28,6 +28,24 @@ def longest_common_subsequence(x: str, y: str):
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(2, 'ph')
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>>> longest_common_subsequence("computer", "food")
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(1, 'o')
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>>> longest_common_subsequence("", "abc") # One string is empty
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(0, '')
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>>> longest_common_subsequence("abc", "") # Other string is empty
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(0, '')
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>>> longest_common_subsequence("", "") # Both strings are empty
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(0, '')
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>>> longest_common_subsequence("abc", "def") # No common subsequence
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(0, '')
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>>> longest_common_subsequence("abc", "abc") # Identical strings
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(3, 'abc')
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>>> longest_common_subsequence("a", "a") # Single character match
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(1, 'a')
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>>> longest_common_subsequence("a", "b") # Single character no match
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(0, '')
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>>> longest_common_subsequence("abcdef", "ace") # Interleaved subsequence
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(3, 'ace')
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>>> longest_common_subsequence("ABCD", "ACBD") # No repeated characters
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(3, 'ABD')
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"""
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# find the length of strings
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@ -1,17 +1,35 @@
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"""
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Numerical integration or quadrature for a smooth function f with known values at x_i
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This method is the classical approach of suming 'Equally Spaced Abscissas'
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This method is the classical approach of summing 'Equally Spaced Abscissas'
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method 1:
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"extended trapezoidal rule"
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Method 1:
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"Extended Trapezoidal Rule"
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"""
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def method_1(boundary, steps):
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# "extended trapezoidal rule"
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# int(f) = dx/2 * (f1 + 2f2 + ... + fn)
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"""
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This function implements the extended trapezoidal rule for numerical integration.
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The function f(x) is provided below.
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:param boundary: List containing the lower and upper bounds of integration [a, b]
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:param steps: The number of steps (intervals) used in the approximation
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:return: The numerical approximation of the integral
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>>> abs(method_1([0, 1], 10) - 0.33333) < 0.01
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True
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>>> abs(method_1([0, 1], 100) - 0.33333) < 0.01
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True
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>>> abs(method_1([0, 2], 1000) - 2.66667) < 0.01
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True
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>>> abs(method_1([1, 2], 1000) - 2.33333) < 0.01
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True
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"""
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h = (boundary[1] - boundary[0]) / steps
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a = boundary[0]
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b = boundary[1]
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@ -19,29 +37,67 @@ def method_1(boundary, steps):
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y = 0.0
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y += (h / 2.0) * f(a)
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for i in x_i:
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# print(i)
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y += h * f(i)
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y += (h / 2.0) * f(b)
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return y
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def make_points(a, b, h):
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"""
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Generates the points between a and b with spacing h for trapezoidal integration.
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:param a: The lower bound of integration
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:param b: The upper bound of integration
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:param h: The step size
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:yield: The next x-value in the range (a, b)
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>>> list(make_points(0, 1, 0.1))
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[0.1, 0.2, 0.30000000000000004, \
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0.4, 0.5, 0.6, 0.7, \
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0.7999999999999999, \
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0.8999999999999999]
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"""
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x = a + h
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while x < (b - h):
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yield x
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x = x + h
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def f(x): # enter your function here
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y = (x - 0) * (x - 0)
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def f(x):
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"""
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This is the function to integrate, f(x) = (x - 0)^2 = x^2.
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:param x: The input value
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:return: The value of f(x)
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>>> f(0)
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0.0
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>>> f(1)
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1.0
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>>> f(0.5)
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0.25
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"""
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y = float((x - 0) * (x - 0))
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return y
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def main():
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a = 0.0 # Lower bound of integration
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b = 1.0 # Upper bound of integration
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steps = 10.0 # define number of steps or resolution
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boundary = [a, b] # define boundary of integration
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"""
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Main function to test the trapezoidal rule.
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:a: Lower bound of integration
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:b: Upper bound of integration
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:steps: define number of steps or resolution
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:boundary: define boundary of integration
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>>> main()
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y = 0.3349999999999999
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"""
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a = 0.0
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b = 1.0
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steps = 10.0
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boundary = [a, b]
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y = method_1(boundary, steps)
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print(f"y = {y}")
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113
searches/exponential_search.py
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113
searches/exponential_search.py
Normal file
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@ -0,0 +1,113 @@
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#!/usr/bin/env python3
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"""
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Pure Python implementation of exponential search algorithm
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For more information, see the Wikipedia page:
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https://en.wikipedia.org/wiki/Exponential_search
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For doctests run the following command:
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python3 -m doctest -v exponential_search.py
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For manual testing run:
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python3 exponential_search.py
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"""
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from __future__ import annotations
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def binary_search_by_recursion(
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sorted_collection: list[int], item: int, left: int = 0, right: int = -1
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) -> int:
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"""Pure implementation of binary search algorithm in Python using recursion
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Be careful: the collection must be ascending sorted otherwise, the result will be
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unpredictable.
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:param sorted_collection: some ascending sorted collection with comparable items
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:param item: item value to search
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:param left: starting index for the search
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:param right: ending index for the search
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:return: index of the found item or -1 if the item is not found
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Examples:
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>>> binary_search_by_recursion([0, 5, 7, 10, 15], 0, 0, 4)
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0
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>>> binary_search_by_recursion([0, 5, 7, 10, 15], 15, 0, 4)
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4
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>>> binary_search_by_recursion([0, 5, 7, 10, 15], 5, 0, 4)
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1
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>>> binary_search_by_recursion([0, 5, 7, 10, 15], 6, 0, 4)
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-1
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"""
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if right < 0:
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right = len(sorted_collection) - 1
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if list(sorted_collection) != sorted(sorted_collection):
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raise ValueError("sorted_collection must be sorted in ascending order")
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if right < left:
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return -1
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midpoint = left + (right - left) // 2
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if sorted_collection[midpoint] == item:
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return midpoint
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elif sorted_collection[midpoint] > item:
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return binary_search_by_recursion(sorted_collection, item, left, midpoint - 1)
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else:
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return binary_search_by_recursion(sorted_collection, item, midpoint + 1, right)
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def exponential_search(sorted_collection: list[int], item: int) -> int:
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"""
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Pure implementation of an exponential search algorithm in Python.
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For more information, refer to:
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https://en.wikipedia.org/wiki/Exponential_search
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Be careful: the collection must be ascending sorted, otherwise the result will be
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unpredictable.
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:param sorted_collection: some ascending sorted collection with comparable items
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:param item: item value to search
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:return: index of the found item or -1 if the item is not found
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The time complexity of this algorithm is O(log i) where i is the index of the item.
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Examples:
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>>> exponential_search([0, 5, 7, 10, 15], 0)
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0
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>>> exponential_search([0, 5, 7, 10, 15], 15)
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4
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>>> exponential_search([0, 5, 7, 10, 15], 5)
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1
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>>> exponential_search([0, 5, 7, 10, 15], 6)
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-1
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"""
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if list(sorted_collection) != sorted(sorted_collection):
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raise ValueError("sorted_collection must be sorted in ascending order")
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if sorted_collection[0] == item:
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return 0
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bound = 1
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while bound < len(sorted_collection) and sorted_collection[bound] < item:
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bound *= 2
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left = bound // 2
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right = min(bound, len(sorted_collection) - 1)
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return binary_search_by_recursion(sorted_collection, item, left, right)
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if __name__ == "__main__":
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import doctest
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doctest.testmod()
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# Manual testing
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user_input = input("Enter numbers separated by commas: ").strip()
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collection = sorted(int(item) for item in user_input.split(","))
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target = int(input("Enter a number to search for: "))
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result = exponential_search(sorted_collection=collection, item=target)
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if result == -1:
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print(f"{target} was not found in {collection}.")
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else:
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print(f"{target} was found at index {result} in {collection}.")
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