Binary Search
Given a sorted array arr[] of n elements, write a function to search a given element x in arr[].
A simple approach is to do linear search.The time complexity of above algorithm is O(n). Another approach to perform the same task is using Binary Search.
Binary Search: Search a sorted array by repeatedly dividing the search interval in half. Begin with an interval covering the whole array. If the value of the search key is less than the item in the middle of the interval, narrow the interval to the lower half. Otherwise narrow it to the upper half. Repeatedly check until the value is found or the interval is empty.
The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(Logn).
We basically ignore half of the elements just after one comparison.
 Compare x with the middle element.
 If x matches with middle element, we return the mid index.
 Else If x is greater than the mid element, then x can only lie in right half subarray after the mid element. So we recur for right half.
 Else (x is smaller) recur for the left half.
Recursive implementation of Binary Search
 C/C++

#include <stdio.h>
// A recursive binary search function. It returns location of x in
// given array arr[l..r] is present, otherwise 1
int
binarySearch(
int
arr[],
int
l,
int
r,
int
x)
{
if
(r >= l)
{
int
mid = l + (r  l)/2;
// If the element is present at the middle itself
if
(arr[mid] == x)
return
mid;
// If element is smaller than mid, then it can only be present
// in left subarray
if
(arr[mid] > x)
return
binarySearch(arr, l, mid1, x);
// Else the element can only be present in right subarray
return
binarySearch(arr, mid+1, r, x);
}
// We reach here when element is not present in array
return
1;
}
int
main(
void
)
{
int
arr[] = {2, 3, 4, 10, 40};
int
n =
sizeof
(arr)/
sizeof
(arr[0]);
int
x = 10;
int
result = binarySearch(arr, 0, n1, x);
(result == 1)?
printf
(
"Element is not present in array"
)
:
printf
(
"Element is present at index %d"
, result);
return
0;
}
Output:
Element is present at index 3
Iterative implementation of Binary Search
 C/C++

#include <stdio.h>
// A iterative binary search function. It returns location of x in
// given array arr[l..r] if present, otherwise 1
int
binarySearch(
int
arr[],
int
l,
int
r,
int
x)
{
while
(l <= r)
{
int
m = l + (rl)/2;
// Check if x is present at mid
if
(arr[m] == x)
return
m;
// If x greater, ignore left half
if
(arr[m] < x)
l = m + 1;
// If x is smaller, ignore right half
else
r = m  1;
}
// if we reach here, then element was not present
return
1;
}
int
main(
void
)
{
int
arr[] = {2, 3, 4, 10, 40};
int
n =
sizeof
(arr)/
sizeof
(arr[0]);
int
x = 10;
int
result = binarySearch(arr, 0, n1, x);
(result == 1)?
printf
(
"Element is not present in array"
)
:
printf
(
"Element is present at index %d"
, result);
return
0;
}
Output:
Element is present at index 3
Time Complexity:
The time complexity of Binary Search can be written as
T(n) = T(n/2) + c
The above recurrence can be solved either using Recurrence T ree method or Master method. It falls in case II of Master Method and solution of the recurrence is .
Auxiliary Space: O(1) in case of iterative implementation. In case of recursive implementation, O(Logn) recursion call stack space.
Algorithmic Paradigm: Divide and Conquer
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source and credits.
source and credits: http://www.geeksforgeeks.org
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