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  • Difficulty Level : Easy
  • Last Updated : 16 Aug, 2022

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 the array.

                 Consider array is 0 base index.

Examples: 

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

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.  

Binary Search Approach: 

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). 

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

  • Begin with the mid element of the whole array as a search key.
  • If the value of the search key is equal to the item then return an 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.

Binary Search Algorithm can be implemented in the following two ways

  1. Iterative Method
  2. Recursive Method

1. Iteration Method

    binarySearch(arr, x, low, high)
        repeat till low = high
               mid = (low + high)/2
                   if (x == arr[mid])
                   return mid
   
                   else if (x > arr[mid]) // x is on the right side
                       low = mid + 1
   
                   else                  // x is on the left side
                       high = mid - 1

2. Recursive Method (The recursive method follows the divide and conquers approach)

    binarySearch(arr, x, low, high)
           if low > high
               return False 
   
           else
               mid = (low + high) / 2 
                   if x == arr[mid]
                   return mid
       
               else if x > arr[mid]        // x is on the right side
                   return binarySearch(arr, x, mid + 1, high)
               
               else                        // x is on the right side
                   return binarySearch(arr, x, low, mid - 1) 

Illustration of Binary Search Algorithm: 

Example of Binary Search Algorithm

Recommended Practice
 

Complete Interview Preparation - GFG

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 <bits/stdc++.h>
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




// C program to implement recursive Binary Search
#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, 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




// 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




# 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#




// 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.

PHP




<?php
// PHP program to implement
// recursive Binary Search
 
// A recursive binary search
// function. It returns location
// of x in given array arr[l..r]
// is present, otherwise -1
function binarySearch($arr, $l, $r, $x)
{
if ($r >= $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.
?>

Javascript




<script>
// JavaScript program to implement recursive Binary Search
 
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
function binarySearch(arr, l, r, x){
    if (r >= l) {
        let mid = l + Math.floor((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;
}
 
let arr = [ 2, 3, 4, 10, 40 ];
let x = 10;
let n = arr.length
let result = binarySearch(arr, 0, n - 1, x);
(result == -1) ? document.write( "Element is not present in array")
                   : document.write("Element is present at index " +result);
</script>

Output

Element is present at index 3

Time Complexity: O(log n)
Auxiliary Space: O(log n)

Another Iterative Approach to Binary Search

C++




#include <bits/stdc++.h>
#include <iostream>
using namespace std;
 
int binarySearch(vector<int> v, int To_Find)
{
    int lo = 0, hi = v.size() - 1;
    int mid;
    // This below check covers all cases , so need to check
    // for mid=lo-(hi-lo)/2
    while (hi - lo > 1) {
        int mid = (hi + lo) / 2;
        if (v[mid] < To_Find) {
            lo = mid + 1;
        }
        else {
            hi = mid;
        }
    }
    if (v[lo] == To_Find) {
        cout << "Found"
             << " At Index " << lo << endl;
    }
    else if (v[hi] == To_Find) {
        cout << "Found"
             << " At Index " << hi << endl;
    }
    else {
        cout << "Not Found" << endl;
    }
}
 
int main()
{
    vector<int> v = { 1, 3, 4, 5, 6 };
    int To_Find = 1;
    binarySearch(v, To_Find);
    To_Find = 6;
    binarySearch(v, To_Find);
    To_Find = 10;
    binarySearch(v, To_Find);
    return 0;
}

Java




/*package whatever //do not write package name here */
 
import java.io.*;
import java.util.*;
 
class GFG {
   
  static void binarySearch(int v[], int To_Find)
{
    int lo = 0, hi = v.length - 1;
    // This below check covers all cases , so need to check
    // for mid=lo-(hi-lo)/2
    while (hi - lo > 1) {
        int mid = (hi + lo) / 2;
        if (v[mid] < To_Find) {
            lo = mid + 1;
        }
        else {
            hi = mid;
        }
    }
    if (v[lo] == To_Find) {
      System.out.println("Found At Index " + lo );
    }
    else if (v[hi] == To_Find) {
        System.out.println("Found At Index " + hi );
    }
    else {
        System.out.println("Not Found" );
    }
}
   
   
    public static void main (String[] args) {
         
      int v[]={1, 3, 4, 5, 6};
       
      /* List<ArrayList<Integer>> v = new ArrayList<>();
      v.add(new ArrayList<Integer>(Arrays.asList( 1, 3, 4, 5, 6 )));*/
    int To_Find = 1;
    binarySearch(v, To_Find);
    To_Find = 6;
    binarySearch(v, To_Find);
    To_Find = 10;
    binarySearch(v, To_Find);
    }
}
// contributed by akashish__

Python3




def binarySearch(v, To_Find):
    lo = 0
    hi = len(v) - 1
 
    # This below check covers all cases , so need to check
    # for mid=lo-(hi-lo)/2
    while hi - lo > 1:
        mid = (hi + lo) // 2
        if v[mid] < To_Find:
            lo = mid + 1
        else:
            hi = mid
 
    if v[lo] == To_Find:
        print("Found At Index", lo)
    elif v[hi] == To_Find:
        print("Found At Index", hi)
    else:
        print("Not Found")
 
 
if __name__ == '__main__':
    v = [1, 3, 4, 5, 6]
 
    To_Find = 1
    binarySearch(v, To_Find)
 
    To_Find = 6
    binarySearch(v, To_Find)
 
    To_Find = 10
    binarySearch(v, To_Find)
 
# This code is contrbuted by Tapesh(tapeshdua420)

C#




using System;
 
public class GFG{
   
  static public void binarySearch(int[] v, int To_Find)
{
    int lo = 0;
    int hi = v.Length - 1;
    int mid;
    // This below check covers all cases , so need to check
    // for mid=lo-(hi-lo)/2
    while (hi - lo > 1) {
        mid = (hi + lo) / 2;
        if (v[mid] < To_Find) {
            lo = mid + 1;
        }
        else {
            hi = mid;
        }
    }
    if (v[lo] == To_Find) {
        Console.WriteLine("Found At Index " + lo);
    }
    else if (v[hi] == To_Find) {
        Console.WriteLine("Found At Index " +  hi);
    }
    else {
       Console.WriteLine("Not Found");
    }
}
 
    static public void Main (){
 
    int[] v = { 1, 3, 4, 5, 6 };
    int To_Find = 1;
    binarySearch(v, To_Find);
    To_Find = 6;
    binarySearch(v, To_Find);
    To_Find = 10;
    binarySearch(v, To_Find);
    }
}
 
// contributed by akashish__

Javascript




<script>
function binarySearch(v,To_Find)
{
    let lo = 0;
    let hi = v.length - 1;
    let mid;
     
    // This below check covers all cases , so need to check
    // for mid=lo-(hi-lo)/2
    while (hi - lo > 1) {
        let mid = (hi + lo) / 2;
        if (v[mid] < To_Find) {
            lo = mid + 1;
        }
        else {
            hi = mid;
        }
    }
    if (v[lo] == To_Find) {
        console.log( "Found At Index " + lo);
    }
    else if (v[hi] == To_Find) {
        console.log("Found At Index " + hi);
    }
    else {
        console.log("Not Found");
    }
}
v = [ 1, 3, 4, 5, 6 ];
let To_Find = 1;
binarySearch(v, To_Find);
To_Find = 6;
binarySearch(v, To_Find);
To_Find = 10;
binarySearch(v, To_Find);
 
// This code is contributed by akashish__
</script>

Output

Found At Index 0
Found At Index 4
Not Found

Time Complexity: O (log n)
Auxiliary Space: O (1)

Iterative implementation of Binary Search 
 

C++




// C++ program to implement iterative Binary Search
#include <bits/stdc++.h>
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




// C program to implement iterative Binary Search
#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 + (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




// 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




# 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#




// 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

PHP




<?php
// PHP program to implement
// iterative Binary Search
 
// A iterative binary search
// function. It returns location
// of x in given array arr[l..r]
// if present, otherwise -1
function binarySearch($arr, $l,
                      $r, $x)
{
    while ($l <= $r)
    {
        $m = $l + ($r - $l) / 2;
 
        // Check if x is present at mid
        if ($arr[$m] == $x)
            return floor($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
$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.
?>

Javascript




<script>
// Program to implement iterative Binary Search
 
  
// A iterative binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
 
 function binarySearch(arr, x)
{   
    let l = 0;
    let r = arr.length - 1;
    let mid;
    while (r >= l) {
         mid = l + Math.floor((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)
            r = mid - 1;
             
        // Else the element can only be present
        // in right subarray
        else
            l = mid + 1;
    }
  
    // We reach here when element is not
    // present in array
    return -1;
}
 
    arr =new Array(2, 3, 4, 10, 40);
    x = 10;
    n = arr.length;
    result = binarySearch(arr, x);
     
(result == -1) ? document.write("Element is not present in array")
               : document.write ("Element is present at index " + result);
                
// This code is contributed by simranarora5sos and rshuklabbb
</script>

Output

Element is present at index 3

Time Complexity: O(log n)
Auxiliary Space: O(1)

Algorithmic Paradigm: Decrease and Conquer.

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.

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