# 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 a linear search. The time complexity of the 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.

Example :

The idea of binary search is to use the information that the array is sorted and reduce the time complexity to O(Log n).

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++ 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;` `}`   `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;` `}`

 `// C program to implement recursive Binary Search` `#include `   `// 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;` `}`   `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, n - 1, x);` `    ``(result == -1) ? ``printf``(``"Element is not present in array"``)` `                   ``: ``printf``(``"Element is present at index %d"``,` `                            ``result);` `    ``return` `0;` `}`

 `// Java implementation of recursive Binary Search` `class` `BinarySearch {` `    ``// Returns index of x if it is present in arr[l..` `    ``// r], else return -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 method to test above` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``BinarySearch ob = ``new` `BinarySearch();` `        ``int` `arr[] = { ``2``, ``3``, ``4``, ``10``, ``40` `};` `        ``int` `n = arr.length;` `        ``int` `x = ``10``;` `        ``int` `result = ob.binarySearch(arr, ``0``, n - ``1``, x);` `        ``if` `(result == -``1``)` `            ``System.out.println(``"Element not present"``);` `        ``else` `            ``System.out.println(``"Element found at index "` `+ result);` `    ``}` `}` `/* This code is contributed by Rajat Mishra */`

 `# Python3 Program for recursive binary search.`   `# Returns index of x in arr if present, else -1` `def` `binarySearch (arr, l, r, x):`   `    ``# Check base case` `    ``if` `r >``=` `l:`   `        ``mid ``=` `l ``+` `(r ``-` `l) ``/``/` `2`   `        ``# If 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` `        ``elif` `arr[mid] > x:` `            ``return` `binarySearch(arr, l, mid``-``1``, x)`   `        ``# Else the element can only be present ` `        ``# in right subarray` `        ``else``:` `            ``return` `binarySearch(arr, mid ``+` `1``, r, x)`   `    ``else``:` `        ``# Element is not present in the array` `        ``return` `-``1`   `# Driver Code` `arr ``=` `[ ``2``, ``3``, ``4``, ``10``, ``40` `]` `x ``=` `10`   `# Function call` `result ``=` `binarySearch(arr, ``0``, ``len``(arr)``-``1``, x)`   `if` `result !``=` `-``1``:` `    ``print` `(``"Element is present at index % d"` `%` `result)` `else``:` `    ``print` `(``"Element is not present in array"``)`

 `// C# implementation of recursive Binary Search` `using` `System;`   `class` `GFG {` `    ``// Returns index of x if it is present in` `    ``// arr[l..r], else return -1` `    ``static` `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 method to test above` `    ``public` `static` `void` `Main()` `    ``{`   `        ``int``[] arr = { 2, 3, 4, 10, 40 };` `        ``int` `n = arr.Length;` `        ``int` `x = 10;`   `        ``int` `result = binarySearch(arr, 0, n - 1, x);`   `        ``if` `(result == -1)` `            ``Console.WriteLine(``"Element not present"``);` `        ``else` `            ``Console.WriteLine(``"Element found at index "` `                              ``+ result);` `    ``}` `}`   `// This code is contributed by Sam007.`

 `= ``\$l``)` `{` `        ``\$mid` `= ``ceil``(``\$l` `+ (``\$r` `- ``\$l``) / 2);`   `        ``// If the element is present ` `        ``// at the middle itself` `        ``if` `(``\$arr``[``\$mid``] == ``\$x``) ` `            ``return` `floor``(``\$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` `\$arr` `= ``array``(2, 3, 4, 10, 40);` `\$n` `= ``count``(``\$arr``);` `\$x` `= 10;` `\$result` `= binarySearch(``\$arr``, 0, ``\$n` `- 1, ``\$x``);` `if``((``\$result` `== -1))` `echo` `"Element is not present in array"``;` `else` `echo` `"Element is present at index "``,` `                            ``\$result``;` `                            `  `// This code is contributed by anuj_67.` `?>`

 ``

Output :

`Element is present at index 3`

Here you can create a check function for easier implementation.

Here is recursive implementation with check function which I feel is a much easier implementation:

 `C++`   `#include ` `using` `namespace` `std;`   `//define array globally` `const` `int` `N = 1e6 +4;`   `int` `a[N];` `int` `n;``//array size`   `//elememt to be searched in array` `   ``int` `k;`   `bool` `check(``int` `dig)` `{` `    ``//element at dig position in array` `    ``int` `ele=a[dig];`   `    ``//if k is less than` `    ``//element at dig position ` `    ``//then we need to bring our higher ending to dig` `    ``//and then continue further` `    ``if``(k<=ele)` `    ``{` `        ``return` `1;` `    ``}` `    ``else` `    ``{` `    ``return` `0;` `    ``}` `}` `void` `binsrch(``int` `lo,``int` `hi)` `{` `    ``while``(lo>n;` `   ``for``(``int` `i=0; i>a[i];` `   ``}` `   ``cin>>k;`   `   ``//it is being given array is sorted` `   ``//if not then we have to sort it`   `   ``//minimum possible point where our k can be is starting index` `   ``//so lo=0 ` `   ``//also k cannot be outside of array so end point` `   ``//hi=n`   `   ``binsrch(0,n);`   `    ``return` `0;` `}`

Iterative implementation of Binary Search

 `// C++ program to implement recursive 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;` `}`   `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;` `}`

 `// C program to implement iterative Binary Search` `#include `   `// 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;` `}`   `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, n - 1, x);` `    ``(result == -1) ? ``printf``(``"Element is not present"` `                            ``" in array"``)` `                   ``: ``printf``(``"Element is present at "` `                            ``"index %d"``,` `                            ``result);` `    ``return` `0;` `}`

 `// Java implementation of iterative Binary Search` `class` `BinarySearch {` `    ``// Returns index of x if it is present in arr[],` `    ``// else return -1` `    ``int` `binarySearch(``int` `arr[], ``int` `x)` `    ``{` `        ``int` `l = ``0``, r = arr.length - ``1``;` `        ``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 method to test above` `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``BinarySearch ob = ``new` `BinarySearch();` `        ``int` `arr[] = { ``2``, ``3``, ``4``, ``10``, ``40` `};` `        ``int` `n = arr.length;` `        ``int` `x = ``10``;` `        ``int` `result = ob.binarySearch(arr, x);` `        ``if` `(result == -``1``)` `            ``System.out.println(``"Element not present"``);` `        ``else` `            ``System.out.println(``"Element found at "` `                               ``+ ``"index "` `+ result);` `    ``}` `}`

 `# Python3 code to implement iterative Binary ` `# Search.`   `# It returns location of x in given array arr ` `# if present, else returns -1` `def` `binarySearch(arr, l, r, x):`   `    ``while` `l <``=` `r:`   `        ``mid ``=` `l ``+` `(r ``-` `l) ``/``/` `2``;` `        `  `        ``# Check if x is present at mid` `        ``if` `arr[mid] ``=``=` `x:` `            ``return` `mid`   `        ``# If x is greater, ignore left half` `        ``elif` `arr[mid] < x:` `            ``l ``=` `mid ``+` `1`   `        ``# If x is smaller, ignore right half` `        ``else``:` `            ``r ``=` `mid ``-` `1` `    `  `    ``# If we reach here, then the element` `    ``# was not present` `    ``return` `-``1`   `# Driver Code` `arr ``=` `[ ``2``, ``3``, ``4``, ``10``, ``40` `]` `x ``=` `10`   `# Function call` `result ``=` `binarySearch(arr, ``0``, ``len``(arr)``-``1``, x)`   `if` `result !``=` `-``1``:` `    ``print` `(``"Element is present at index % d"` `%` `result)` `else``:` `    ``print` `(``"Element is not present in array"``)`

 `// C# implementation of iterative Binary Search` `using` `System;`   `class` `GFG {` `    ``// Returns index of x if it is present in arr[],` `    ``// else return -1` `    ``static` `int` `binarySearch(``int``[] arr, ``int` `x)` `    ``{` `        ``int` `l = 0, r = arr.Length - 1;` `        ``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 method to test above` `    ``public` `static` `void` `Main()` `    ``{` `        ``int``[] arr = { 2, 3, 4, 10, 40 };` `        ``int` `n = arr.Length;` `        ``int` `x = 10;` `        ``int` `result = binarySearch(arr, x);` `        ``if` `(result == -1)` `            ``Console.WriteLine(``"Element not present"``);` `        ``else` `            ``Console.WriteLine(``"Element found at "` `                              ``+ ``"index "` `+ result);` `    ``}` `}` `// This code is contributed by Sam007`

 ``

 ``

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 the Recurrence Tree method or Master method. It falls in case II of the Master Method and the solution of the recurrence is .
Auxiliary Space: O(1) in case of iterative implementation. In the case of recursive implementation, O(Logn) recursion call stack space.

Note:

Here we are using

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. 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.