 # Binary Search

## Binary Search

Binary Search is a searching algorithm for searching an element in a sorted list or array. This Algorithm is efficient than the Linear Search algorithm and performs the search operation in logarithmic time complexity for sorted arrays or lists.

Binary Search performs the search operation by repeatedly dividing the search interval in half. The idea is, to 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. continuously, check until the value is found or the interval is empty.

Problem: Given a sorted array arr[] of N elements, write a function to search a given element X in arr[] using Binary Search Algorithm

#### Algorithm

• Compare X with the middle element of the array.
• 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 the right half subarray after the mid element. So we will now look for X in only the right half ignoring the complete left half.
• Else if X is smaller, search for X in the left half ignoring the right half.

#### Implementation

``````
// A recursive binary search function. It returns
// location of x in given array arr[l..r] if present,
// otherwise -1

// Initially,
// l = 0, first index of arr[].
// r = N-1, last index of arr[].
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, mid - 1, 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;
}
``````

An Iterative Binary Search Approach

``````
// A iterative binary search function. It returns
// location of x in given array arr[l..r] if present,
// otherwise -1

// Initially,
// l = 0, first index of arr[].
// r = N-1, last index of arr[].
int binarySearch(int arr[], int l, int r, int x)
{
while (l <= r) {
int m = l + (r - l) / 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;
}
``````

#### Time Complexity

O(log N), where N is the number of elements in the array Sudhanshu is Technology geek and also a pro pubg player. I like to create content in my free time that helps others. Currently pursuing BCA from Noida and operating Geekstocode

## 1 Comment

1. Kristian says:

Nicely done & written mmy friend!
I started blogging myself recently and have seen that
lot of bloggers simply rehas oldd ideas but add very little of worth.
It’s good to see an educational article of some adtual value too me and