#!/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
"""
import bisect
from typing import List, Optional
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) // 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) // 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) -> Optional[int]:
"""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) -> Optional[int]:
"""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
) -> Optional[int]:
"""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:
print(f"{target} was not found in {collection}.")
else:
print(f"{target} was found at position {result} in {collection}.")
Given a sorted array of n elements, write a function to search for the index of a given element (target)
O(log n) Worse Case
O(1) Best Case (If middle element of initial array is the target element)
O(1) For iterative approach
O(log n) For recursive approach due to recursion call stack
arr = [1,2,3,4,5,6,7]
target = 2
Initially the element at middle index is 4 which is greater than 2. Therefore we search the left half of the
array i.e. [1,2,3].
Here we find the middle element equal to target element so we return its index i.e. 1
target = 9
Binary Search should return -1 as 9 is not present in the array
A CS50 video explaining the Binary Search Algorithm