#### Binary Search

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#!/usr/bin/env python3

"""
This is pure Python implementation of binary search algorithms

For doctests run following command:
python3 -m doctest -v binary_search.py

For manual testing run:
python3 binary_search.py
"""
from __future__ import annotations

import bisect

def bisect_left(
sorted_collection: list[int], item: int, lo: int = 0, hi: int = -1
) -> int:
"""
Locates the first element in a sorted array that is larger or equal to a given
value.

It has the same interface as
https://docs.python.org/3/library/bisect.html#bisect.bisect_left .

:param sorted_collection: some ascending sorted collection with comparable items
:param item: item to bisect
:param lo: lowest index to consider (as in sorted_collection[lo:hi])
:param hi: past the highest index to consider (as in sorted_collection[lo:hi])
:return: index i such that all values in sorted_collection[lo:i] are < item and all
values in sorted_collection[i:hi] are >= item.

Examples:
>>> bisect_left([0, 5, 7, 10, 15], 0)
0

>>> bisect_left([0, 5, 7, 10, 15], 6)
2

>>> bisect_left([0, 5, 7, 10, 15], 20)
5

>>> bisect_left([0, 5, 7, 10, 15], 15, 1, 3)
3

>>> bisect_left([0, 5, 7, 10, 15], 6, 2)
2
"""
if hi < 0:
hi = len(sorted_collection)

while lo < hi:
mid = lo + (hi - lo) // 2
if sorted_collection[mid] < item:
lo = mid + 1
else:
hi = mid

return lo

def bisect_right(
sorted_collection: list[int], item: int, lo: int = 0, hi: int = -1
) -> int:
"""
Locates the first element in a sorted array that is larger than a given value.

It has the same interface as
https://docs.python.org/3/library/bisect.html#bisect.bisect_right .

:param sorted_collection: some ascending sorted collection with comparable items
:param item: item to bisect
:param lo: lowest index to consider (as in sorted_collection[lo:hi])
:param hi: past the highest index to consider (as in sorted_collection[lo:hi])
:return: index i such that all values in sorted_collection[lo:i] are <= item and
all values in sorted_collection[i:hi] are > item.

Examples:
>>> bisect_right([0, 5, 7, 10, 15], 0)
1

>>> bisect_right([0, 5, 7, 10, 15], 15)
5

>>> bisect_right([0, 5, 7, 10, 15], 6)
2

>>> bisect_right([0, 5, 7, 10, 15], 15, 1, 3)
3

>>> bisect_right([0, 5, 7, 10, 15], 6, 2)
2
"""
if hi < 0:
hi = len(sorted_collection)

while lo < hi:
mid = lo + (hi - lo) // 2
if sorted_collection[mid] <= item:
lo = mid + 1
else:
hi = mid

return lo

def insort_left(
sorted_collection: list[int], item: int, lo: int = 0, hi: int = -1
) -> None:
"""
Inserts a given value into a sorted array before other values with the same value.

It has the same interface as
https://docs.python.org/3/library/bisect.html#bisect.insort_left .

:param sorted_collection: some ascending sorted collection with comparable items
:param item: item to insert
:param lo: lowest index to consider (as in sorted_collection[lo:hi])
:param hi: past the highest index to consider (as in sorted_collection[lo:hi])

Examples:
>>> sorted_collection = [0, 5, 7, 10, 15]
>>> insort_left(sorted_collection, 6)
>>> sorted_collection
[0, 5, 6, 7, 10, 15]

>>> sorted_collection = [(0, 0), (5, 5), (7, 7), (10, 10), (15, 15)]
>>> item = (5, 5)
>>> insort_left(sorted_collection, item)
>>> sorted_collection
[(0, 0), (5, 5), (5, 5), (7, 7), (10, 10), (15, 15)]
>>> item is sorted_collection[1]
True
>>> item is sorted_collection[2]
False

>>> sorted_collection = [0, 5, 7, 10, 15]
>>> insort_left(sorted_collection, 20)
>>> sorted_collection
[0, 5, 7, 10, 15, 20]

>>> sorted_collection = [0, 5, 7, 10, 15]
>>> insort_left(sorted_collection, 15, 1, 3)
>>> sorted_collection
[0, 5, 7, 15, 10, 15]
"""
sorted_collection.insert(bisect_left(sorted_collection, item, lo, hi), item)

def insort_right(
sorted_collection: list[int], item: int, lo: int = 0, hi: int = -1
) -> None:
"""
Inserts a given value into a sorted array after other values with the same value.

It has the same interface as
https://docs.python.org/3/library/bisect.html#bisect.insort_right .

:param sorted_collection: some ascending sorted collection with comparable items
:param item: item to insert
:param lo: lowest index to consider (as in sorted_collection[lo:hi])
:param hi: past the highest index to consider (as in sorted_collection[lo:hi])

Examples:
>>> sorted_collection = [0, 5, 7, 10, 15]
>>> insort_right(sorted_collection, 6)
>>> sorted_collection
[0, 5, 6, 7, 10, 15]

>>> sorted_collection = [(0, 0), (5, 5), (7, 7), (10, 10), (15, 15)]
>>> item = (5, 5)
>>> insort_right(sorted_collection, item)
>>> sorted_collection
[(0, 0), (5, 5), (5, 5), (7, 7), (10, 10), (15, 15)]
>>> item is sorted_collection[1]
False
>>> item is sorted_collection[2]
True

>>> sorted_collection = [0, 5, 7, 10, 15]
>>> insort_right(sorted_collection, 20)
>>> sorted_collection
[0, 5, 7, 10, 15, 20]

>>> sorted_collection = [0, 5, 7, 10, 15]
>>> insort_right(sorted_collection, 15, 1, 3)
>>> sorted_collection
[0, 5, 7, 15, 10, 15]
"""
sorted_collection.insert(bisect_right(sorted_collection, item, lo, hi), item)

def binary_search(sorted_collection: list[int], item: int) -> int | None:
"""Pure implementation of binary search algorithm in Python

Be careful collection must be ascending sorted, otherwise result will be
unpredictable

:param sorted_collection: some ascending sorted collection with comparable items
:param item: item value to search
:return: index of found item or None if item is not found

Examples:
>>> binary_search([0, 5, 7, 10, 15], 0)
0

>>> binary_search([0, 5, 7, 10, 15], 15)
4

>>> binary_search([0, 5, 7, 10, 15], 5)
1

>>> binary_search([0, 5, 7, 10, 15], 6)

"""
left = 0
right = len(sorted_collection) - 1

while left <= right:
midpoint = left + (right - left) // 2
current_item = sorted_collection[midpoint]
if current_item == item:
return midpoint
elif item < current_item:
right = midpoint - 1
else:
left = midpoint + 1
return None

def binary_search_std_lib(sorted_collection: list[int], item: int) -> int | None:
"""Pure implementation of binary search algorithm in Python using stdlib

Be careful collection must be ascending sorted, otherwise result will be
unpredictable

:param sorted_collection: some ascending sorted collection with comparable items
:param item: item value to search
:return: index of found item or None if item is not found

Examples:
>>> binary_search_std_lib([0, 5, 7, 10, 15], 0)
0

>>> binary_search_std_lib([0, 5, 7, 10, 15], 15)
4

>>> binary_search_std_lib([0, 5, 7, 10, 15], 5)
1

>>> binary_search_std_lib([0, 5, 7, 10, 15], 6)

"""
index = bisect.bisect_left(sorted_collection, item)
if index != len(sorted_collection) and sorted_collection[index] == item:
return index
return None

def binary_search_by_recursion(
sorted_collection: list[int], item: int, left: int, right: int
) -> int | None:

"""Pure implementation of binary search algorithm in Python by recursion

Be careful collection must be ascending sorted, otherwise result will be
unpredictable
First recursion should be started with left=0 and right=(len(sorted_collection)-1)

:param sorted_collection: some ascending sorted collection with comparable items
:param item: item value to search
:return: index of found item or None if 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)

"""
if right < left:
return None

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)

if __name__ == "__main__":
user_input = input("Enter numbers separated by comma:\n").strip()
collection = sorted(int(item) for item in user_input.split(","))
target = int(input("Enter a single number to be found in the list:\n"))
result = binary_search(collection, target)
if result is None:
else:
print(f"{target} was found at position {result} in {collection}.")

#### Declaración de problema

Dada una matriz ordenada de n elementos, escriba una función para buscar el índice de un elemento determinado (destino).

#### Enfoque

• Se busca la matriz dividiendo la matriz por la mitad repetidamente.
• Inicialmente, se considera la matriz real y se selecciona el elemento en el índice medio.
• Se mantiene el índice más bajo, el número 0, y el más alto, la longitud de la matriz.
• Si es igual al elemento de destino, se devuelve el índice.
• De lo contrario, si es mayor que el elemento de destino, se condiera únicamente la mitad izquierda de la matriz (índice inferior = 0, superior = medio - 1).
• De lo contrario, si es menor que el elemento de destino, se considera únicamente la mitad derecha de la matriz (índice inferior = medio + 1, más alto = longitud de la matriz).
• Se devuelve -1 si el elemento de destino no se encuentra en la matriz (caso base: si el índice inferior es mayor o igual que el índice superior).

O(log n)- En el peor de los casos O(1)- En el mejor de los casos (Si el elemento central de la matriz inicial es el elemento de destino)

O(1)- Para un enfoque iterativo O(log n)- Para un enfoque recursivo debido a la pila de llamadas de recursividad

#### Ejemplo

arr = [1,2,3,4,5,6,7]

target = 2
Inicialmente, el elemento en el índice medio es 4, que es mayor que 2. Por lo tanto, buscamos la mitad izquierda de la
matriz, es decir: [1,2,3].
Aquí encontramos el elemento central igual al elemento objetivo, por lo que devolvemos su índice: 1

target = 9
Búsqueda binaria debe devolver -1 dado que 9 no está presente en la matriz

#### Explicación en vídeo de YouTube

Un vídeo CS50 explicando el algoritmo de búsqueda binaria