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:
print(f"{target} was not found in {collection}.")
else:
print(f"{target} was found at position {result} in {collection}.")
About this Algorithm
Problem Statement
Given a sorted array of n elements, write a function to search for the index of a given element (target)
Approach
- Search for the array by dividing the array in half repeatedly.
- Initially consider the actual array and pick the element at the middle index
- Keep a lower index i.e. 0 and higher index i.e. length of array
- If it is equal to the target element then return the index
- Else if it is greater than the target element then consider only the left half of array. (lower index = 0, higher = middle - 1)
- Else if it is less than the target element then consider only the right half of array. (lower index = middle + 1, higher = length of array)
- Return -(insertion index + 1) if the target element is not found in the array (If the lower index is greater than or equal to higher index). Some simpler implementations just return -1 if the element is not found. The offset of 1 must be added as the insertion index might be 0 (the searched value might be smaller than all elements in the array). As indexing starts at 0, this must be distinguishable from the case where the target element has the index 0.
Time Complexity
O(log n) Worst Case
O(1) Best Case (If middle element of initial array is the target element)
Space Complexity
O(1) For iterative approach
O(1) For recursive approach if tail call optimization is used, O(log n) due to recursion call stack, otherwise
Example
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
A simple Binary Search implementation may return -1 as 9 is not present in the array. A more complex one would return the index at which 9 would have to be inserted, which would be `-8` (last position in the array (7) plus one (7+1), negated)`.
Code Implementation Links
Video Explanation
A CS50 video explaining the Binary Search Algorithm
