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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
"""
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}.")

Binary Search

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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 -1 if target element is not found in the array (Base Case: If lower index is greater than or equal to higher index)

Time Complexity

O(log n) Worse Case
O(1) Best Case (If middle element of initial array is the target element)

Space Complexity

O(1) For iterative approach
O(log n) For recursive approach due to recursion call stack

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          
Binary Search should return -1 as 9 is not present in the array

Code Implementation Links

Video Explanation

A CS50 video explaining the Binary Search Algorithm

Animation Explanation