Open in App
Not now

# C++ Program For Binary Search

• Last Updated : 17 Jan, 2023

Problem: Given a sorted array arr[] of n elements, write a function to search a given element x in arr[] and return the index of x in array. Consider array is 0 base index.

Examples:Â

Input: arr[] = {10, 20, 30, 50, 60, 80, 110, 130, 140, 170}, x = 110
Output: 6
Explanation: Element x is present at index 6.Â

Input: arr[] = {10, 20, 30, 40, 60, 110, 120, 130, 170}, x = 175
Output: -1
Explanation: Element x is not present in arr[].

Linear Search Approach: A simple approach is to do a linear search. The time complexity of the Linear search is O(n). Another approach to perform the same task is using Binary Search.Â Â

Binary Search Approach:Â

Binary Search is a searching algorithm used in a sorted array by repeatedly dividing the search interval in half. The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(Log n).Â

Binary Search Algorithm: The basic steps to perform Binary Search are:

• Begin with the mid element of the whole array as search key.
• If the value of the search key is equal to the item then return index of the search key.
• Or 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 from the second point until the value is found or the interval is empty.
Â

Illustration of Binary Search Algorithm:Â

Example of Binary Search Algorithm

Step-by-step Binary Search Algorithm: We basically ignore half of the elements just after one comparison.

1. Compare x with the middle element.
2. If x matches with the middle element, we return the mid index.
3. 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 recur for the right half.
4. Else (x is smaller) recur for the left half.

Recursive implementation of Binary Search:

## C++

 `// C++ program to implement recursive  ` `// Binary Search ` `#include ` `using` `namespace` `std; ` ` `  `// 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,  ` `                                ``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; ` `} ` ` `  `// Driver code ` `int` `main(``void``) ` `{ ` `    ``int` `arr[] = {2, 3, 4, 10, 40}; ` `    ``int` `x = 10; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `result = binarySearch(arr, 0,  ` `                              ``n - 1, x); ` `    ``(result == -1) ? cout <<  ` `     ``"Element is not present in array"` `:  ` `      ``cout << ``"Element is present at index "` `<<  ` `               ``result; ` `    ``return` `0; ` `}`

Output

`Element is present at index 3`

Time Complexity : O(log n)

Auxiliary Space : O(log n)

Iterative implementation of Binary SearchÂ
Â

## C++

 `// C++ program to implement  ` `// iterative Binary Search ` `#include ` `using` `namespace` `std; ` ` `  `// 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 + (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; ` `} ` ` `  `// Driver code ` `int` `main(``void``) ` `{ ` `    ``int` `arr[] = {2, 3, 4, 10, 40}; ` `    ``int` `x = 10; ` `    ``int` `n = ``sizeof``(arr) / ``sizeof``(arr[0]); ` `    ``int` `result = binarySearch(arr, 0,  ` `                              ``n - 1, x); ` `    ``(result == -1) ? cout <<  ` `     ``"Element is not present in array"` `:  ` `      ``cout << ``"Element is present at index "` `<<  ` `               ``result; ` `    ``return` `0; ` `}`

Output

`Element is present at index 3`

Time Complexity : O(log n)

Auxiliary Space : O(1)

Note: Here we are usingÂ

int mid = low + (high – low)/2;

Maybe, you wonder why we are calculating the middle index this way, we can simply add the lower and higher index and divide it by 2.

int mid = (low + high)/2;

But if we calculate the middle index like this means our code is not 100% correct, it contains bugs.

That is, it fails for larger values of int variables low and high. Specifically, it fails if the sum of low and high is greater than the maximum positive int value(231 – 1 ).

The sum overflows to a negative value and the value stays negative when divided by 2.Â
In java, it throws ArrayIndexOutOfBoundException.

int mid = low + (high – low)/2;

So it’s better to use it like this. This bug applies equally to merge sort and other divide and conquer algorithms.

My Personal Notes arrow_drop_up
Related Articles