# Binary Search

Binary Search tries to find if an element is present in a sorted array. It compares the middle element of the array with the searched element.
Thus we get the following cases,

Case 1 : If the middle element equals the searched element, the position of the middle element is returned.
Case 2 : If the middle element is bigger than the searched element, the left part of the array is searched using the same logic i.e binary search.
Left part of the array : [ 0 … mid - 1 ]
Case 3 : If the middle element is smaller than the searched element, the right part of the array is searched using the same logic i.e binary search.
Right part of the array : [ mid + 1 … end-1 ]

Algorithm : Binary Search

1. Binary Search ( Array, Search element N )
2.     Beg = 0, End = Sizeof ( Array ) - 1
3.     While (Beg <= End)
4.         Mid = Beg + ( End - Beg ) / 2
5.         If ( Array [ Mid ] == N )
6.             Return Mid as the index of the found element N.
7.         Else If ( Array [ Mid ] < N )
8.             Beg = Mid + 1.
9.         Else
10.            End = Mid - 1.
11.     Return -1 if the element that we are looking for i.e N is not found in the array.

Example Time complexity : Log ( N )

Why is mid calculated as mid = beg + (end-beg)/2 ?

Integer range is -2,147,483,648 to 2,147,483,647. If you are searching in an array of size 2,000,000,000 and the element searched for is located at index 1,999,999,999. When you search in the upper half of array, beg=1,000,000,001 and end=2,000,000,000. If mid is calculated as (low+high)/2, low+high = 3,000,000,001; which exceeds the range of int, resulting in overflow errors. But mid calculated as beg + (end-beg) = 1,000,000,001 + 999,999,999 = 2,000,000,000; which fits in the integer range.

Binary Search Implementation

``````from typing import List # For annotations

def BinarySearch (array : List[int], n : int) :

beg = 0
end = len(array) - 1

while (beg <= end) :

mid = int(beg + (end-beg)/2)

if (array[mid] == n) :
return mid
elif (array[mid] < n) :
beg = mid + 1;
else :
end = mid - 1

return -1;

array = [ 3, 5, 7, 11, 12, 16, 17, 52, 64, 101 ]

print("Array : " + str(array))
n = int(input("Element to search : "))
index = BinarySearch (array, n)

if (index != -1) :
print("Element found at index : " + str(index))
else :
``````

Output

``````Array : [3, 5, 7, 11, 12, 16, 17, 52, 64, 101]
Element to search : 17
Element found at index : 6

Array : [3, 5, 7, 11, 12, 16, 17, 52, 64, 101]
Element to search : 52
Element found at index : 7

Array : [3, 5, 7, 11, 12, 16, 17, 52, 64, 101]
Element to search : 64
Element found at index : 8

Array : [3, 5, 7, 11, 12, 16, 17, 52, 64, 101]
Element to search : 3
Element found at index : 0

Array : [3, 5, 7, 11, 12, 16, 17, 52, 64, 101]
Element to search : 4
``````
``````#include<iostream>
#include<vector>

int BinarySearch (std :: vector<int> &vec, int n) {

int beg = 0;
int end = vec.size() - 1;

while (beg <= end) {

/* Avoid calculating mid as (beg+end)/2. As (beg+end)
might exceed the range of an integer causing overflow
resulting in errors */

int mid = beg + ( end - beg ) / 2;

if (vec[mid] == n)
return mid;
else if (vec[mid] < n)
beg = mid + 1;
else
end = mid - 1;
}

return -1;
}

int main() {

int n, index;
std :: vector<int> vec = { 3, 5, 7, 11, 12, 16, 17, 52, 64, 101 };

std :: cout << "Array elements :";
for (auto& i : vec) {
std :: cout << i << " ";
} std :: cout << std :: endl;

std :: cout << "Element to search : ";
std :: cin >> n;

if((index = BinarySearch (vec, n)) != -1)
std :: cout << "Element found at index : " <<  index << std :: endl;
else
std :: cout << "Element not found. " << std :: endl;
return 0;
}
``````

Output

``````Array elements :3 5 7 11 12 16 17 52 64 101
Element to search : 11
Element found at index : 3

Array elements :3 5 7 11 12 16 17 52 64 101
Element to search : 7
Element found at index : 2

Array elements :3 5 7 11 12 16 17 52 64 101
Element to search : 64
Element found at index : 8

Array elements :3 5 7 11 12 16 17 52 64 101
Element to search : 100