Search an element in a sorted and rotated Array

Given a sorted and rotated array arr[] of size N and a key, the task is to find the key in the array.

Note: Find the element in O(logN) time and assume that all the elements are distinct.

Example:

Input  : arr[] = {5, 6, 7, 8, 9, 10, 1, 2, 3}, key = 3
Output : Found at index 8

Input  : arr[] = {5, 6, 7, 8, 9, 10, 1, 2, 3}, key = 30
Output : Not found

Input : arr[] = {30, 40, 50, 10, 20}, key = 10   
Output : Found at index 3

Approach 1 (Finding Pivot where rotation has happened): The primary idea to solve the problem is as follows. 

The idea is to find the pivot point, divide the array into two sub-arrays and perform a binary search.

The main idea for finding a pivot is – 

• For a sorted (in increasing order) and rotated array, the pivot element is the only element for which the next element to it is smaller than it.
• Using binary search based on the above idea, pivot can be found.
  • It can be observed that for a search space of indices in range [l, r] where the middle index is mid, 
    • If rotation has happened in the left half, then obviously the element at l will be greater than the one at mid.
    • Otherwise the left half will be sorted but the element at mid will be greater than the one at r.
• After the pivot is found divide the array into two sub-arrays.
• Now the individual sub-arrays are sorted so the element can be searched using Binary Search.

Follow the steps mentioned below to implement the idea:  

• Find out the pivot point using binary search. We will set the low pointer as the first array index and high with the last array index.
  • From the high and low we will calculate the mid value. 
  • If the value at mid-1 is greater than the one at mid, return that value as the pivot.
  • Else if the value at the mid+1 is less than mid, return mid value as the pivot.
  • Otherwise, if the value at low position is greater than mid position, consider the left half. Otherwise, consider the right half.
• Divide the array into two sub-arrays based on the pivot that was found.
• Now call binary search for one of the two sub-arrays.
  • If the element is greater than the 0^th element then search in the left array
  • Else search in the right array.
• If the element is found in the selected sub-array then return the index
• Else return -1.

Follow the below illustration for a better understanding

Illustration:

Consider arr[] = {3, 4, 5, 1, 2}, key = 1

Pivot finding:

low = 0, high = 4:
        =>  mid = 2
        =>  arr[mid] = 5, arr[mid + 1] = 1
        => arr[mid] > arr[mid +1],
        => Therefore the pivot = mid = 2

Array is divided into two parts {3, 4, 5}, {1, 2}
Now  according to the conditions and the key, we need to find in the part {1, 2} 

Key Finding:

We will apply Binary search on {1, 2}. 
low = 3 , high = 4.
=>  mid = 3
=>  arr[mid] = 1 , key = 1, hence arr[mid] = key matches.
=>  The required index = mid = 3

So the element is  found at index 3.

Below is the implementation of the above approach:  

C++

// C++ Program to search an element
// in a sorted and pivoted array

#include <bits/stdc++.h>
using namespace std;

// Standard Binary Search function
int binarySearch(int arr[], int low, int high, int key)
{
    if (high < low)
        return -1;

    int mid = (low + high) / 2;
    if (key == arr[mid])
        return mid;

    if (key > arr[mid])
        return binarySearch(arr, (mid + 1), high, key);

    return binarySearch(arr, low, (mid - 1), key);
}

// Function to get pivot. For array 3, 4, 5, 6, 1, 2
// it returns 3 (index of 6)
int findPivot(int arr[], int low, int high)
{
    // Base cases
    if (high < low)
        return -1;
    if (high == low)
        return low;

    // low + (high - low)/2;
    int mid = (low + high) / 2;
    if (mid < high && arr[mid] > arr[mid + 1])
        return mid;

    if (mid > low && arr[mid] < arr[mid - 1])
        return (mid - 1);

    if (arr[low] >= arr[mid])
        return findPivot(arr, low, mid - 1);

    return findPivot(arr, mid + 1, high);
}

// Searches an element key in a pivoted
// sorted array arr[] of size n
int pivotedBinarySearch(int arr[], int n, int key)
{
    int pivot = findPivot(arr, 0, n - 1);

    // If we didn't find a pivot,
    // then array is not rotated at all
    if (pivot == -1)
        return binarySearch(arr, 0, n - 1, key);

    // If we found a pivot, then first compare with pivot
    // and then search in two subarrays around pivot
    if (arr[pivot] == key)
        return pivot;

    if (arr[0] <= key)
        return binarySearch(arr, 0, pivot - 1, key);

    return binarySearch(arr, pivot + 1, n - 1, key);
}

// Driver program to check above functions
int main()
{
    // Let us search 3 in below array
    int arr1[] = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int key = 3;

    // Function calling
    cout << "Index of the element is : "
         << pivotedBinarySearch(arr1, n, key);

    return 0;
}

C

/* Program to search an element in
   a sorted and pivoted array*/
#include <stdio.h>

int findPivot(int[], int, int);
int binarySearch(int[], int, int, int);

/* Searches an element key in a pivoted
   sorted array arrp[] of size n */
int pivotedBinarySearch(int arr[], int n, int key)
{
    int pivot = findPivot(arr, 0, n - 1);

    // If we didn't find a pivot,
    // then array is not rotated at all
    if (pivot == -1)
        return binarySearch(arr, 0, n - 1, key);

    // If we found a pivot, then first
    // compare with pivot and then
    // search in two subarrays around pivot
    if (arr[pivot] == key)
        return pivot;
    if (arr[0] <= key)
        return binarySearch(arr, 0, pivot - 1, key);
    return binarySearch(arr, pivot + 1, n - 1, key);
}

/* Function to get pivot. For array
   3, 4, 5, 6, 1, 2 it returns 3 (index of 6) */
int findPivot(int arr[], int low, int high)
{
    // base cases
    if (high < low)
        return -1;
    if (high == low)
        return low;

    int mid = (low + high) / 2; /*low + (high - low)/2;*/
    if (mid < high && arr[mid] > arr[mid + 1])
        return mid;
    if (mid > low && arr[mid] < arr[mid - 1])
        return (mid - 1);
    if (arr[low] >= arr[mid])
        return findPivot(arr, low, mid - 1);
    return findPivot(arr, mid + 1, high);
}

/* Standard Binary Search function*/
int binarySearch(int arr[], int low, int high, int key)
{
    if (high < low)
        return -1;
    int mid = (low + high) / 2; /*low + (high - low)/2;*/
    if (key == arr[mid])
        return mid;
    if (key > arr[mid])
        return binarySearch(arr, (mid + 1), high, key);
    return binarySearch(arr, low, (mid - 1), key);
}

/* Driver program to check above functions */
int main()
{
    // Let us search 3 in below array
    int arr1[] = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };
    int n = sizeof(arr1) / sizeof(arr1[0]);
    int key = 3;
    printf("Index of the element is : %d",
           pivotedBinarySearch(arr1, n, key));
    return 0;
}

Java

/* Java program to search an element
   in a sorted and pivoted array*/
import java.io.*;
class Main {

    /* Searches an element key in a
       pivoted sorted array arrp[]
       of size n */
    static int pivotedBinarySearch(int arr[], int n,
                                   int key)
    {
        int pivot = findPivot(arr, 0, n - 1);

        // If we didn't find a pivot, then
        // array is not rotated at all
        if (pivot == -1)
            return binarySearch(arr, 0, n - 1, key);

        // If we found a pivot, then first
        // compare with pivot and then
        // search in two subarrays around pivot
        if (arr[pivot] == key)
            return pivot;
        if (arr[0] <= key)
            return binarySearch(arr, 0, pivot - 1, key);
        return binarySearch(arr, pivot + 1, n - 1, key);
    }

    /* Function to get pivot. For array
       3, 4, 5, 6, 1, 2 it returns
       3 (index of 6) */
    static int findPivot(int arr[], int low, int high)
    {
        // base cases
        if (high < low)
            return -1;
        if (high == low)
            return low;

        /* low + (high - low)/2; */
        int mid = (low + high) / 2;
        if (mid < high && arr[mid] > arr[mid + 1])
            return mid;
        if (mid > low && arr[mid] < arr[mid - 1])
            return (mid - 1);
        if (arr[low] >= arr[mid])
            return findPivot(arr, low, mid - 1);
        return findPivot(arr, mid + 1, high);
    }

    /* Standard Binary Search function */
    static int binarySearch(int arr[], int low, int high,
                            int key)
    {
        if (high < low)
            return -1;

        /* low + (high - low)/2; */
        int mid = (low + high) / 2;
        if (key == arr[mid])
            return mid;
        if (key > arr[mid])
            return binarySearch(arr, (mid + 1), high, key);
        return binarySearch(arr, low, (mid - 1), key);
    }

    // main function
    public static void main(String args[])
    {
        // Let us search 3 in below array
        int arr1[] = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };
        int n = arr1.length;
        int key = 3;
        System.out.println(
            "Index of the element is : "
            + pivotedBinarySearch(arr1, n, key));
    }
}

Python3

# Python Program to search an element
# in a sorted and pivoted array

# Searches an element key in a pivoted
# sorted array arrp[] of size n
def pivotedBinarySearch(arr, n, key):

    pivot = findPivot(arr, 0, n-1)

    # If we didn't find a pivot,
    # then array is not rotated at all
    if pivot == -1:
        return binarySearch(arr, 0, n-1, key)

    # If we found a pivot, then first
    # compare with pivot and then
    # search in two subarrays around pivot
    if arr[pivot] == key:
        return pivot
    if arr[0] <= key:
        return binarySearch(arr, 0, pivot-1, key)
    return binarySearch(arr, pivot + 1, n-1, key)


# Function to get pivot. For array
# 3, 4, 5, 6, 1, 2 it returns 3
# (index of 6)
def findPivot(arr, low, high):

    # base cases
    if high < low:
        return -1
    if high == low:
        return low

    # low + (high - low)/2;
    mid = int((low + high)/2)

    if mid < high and arr[mid] > arr[mid + 1]:
        return mid
    if mid > low and arr[mid] < arr[mid - 1]:
        return (mid-1)
    if arr[low] >= arr[mid]:
        return findPivot(arr, low, mid-1)
    return findPivot(arr, mid + 1, high)

# Standard Binary Search function
def binarySearch(arr, low, high, key):

    if high < low:
        return -1

    # low + (high - low)/2;
    mid = int((low + high)/2)

    if key == arr[mid]:
        return mid
    if key > arr[mid]:
        return binarySearch(arr, (mid + 1), high,
                            key)
    return binarySearch(arr, low, (mid - 1), key)


# Driver program to check above functions
# Let us search 3 in below array
if __name__ == '__main__':
    arr1 = [5, 6, 7, 8, 9, 10, 1, 2, 3]
    n = len(arr1)
    key = 3
    print("Index of the element is : ", \
          pivotedBinarySearch(arr1, n, key))

# This is contributed by Smitha Dinesh Semwal

C#

// C# program to search an element
// in a sorted and pivoted array
using System;

class main {

    // Searches an element key in a
    // pivoted sorted array arrp[]
    // of size n
    static int pivotedBinarySearch(int[] arr,
                                   int n, int key)
    {
        int pivot = findPivot(arr, 0, n - 1);

        // If we didn't find a pivot, then
        // array is not rotated at all
        if (pivot == -1)
            return binarySearch(arr, 0, n - 1, key);

        // If we found a pivot, then first
        // compare with pivot and then
        // search in two subarrays around pivot
        if (arr[pivot] == key)
            return pivot;

        if (arr[0] <= key)
            return binarySearch(arr, 0, pivot - 1, key);

        return binarySearch(arr, pivot + 1, n - 1, key);
    }

    /* Function to get pivot. For array
    3, 4, 5, 6, 1, 2 it returns
    3 (index of 6) */
    static int findPivot(int[] arr, int low, int high)
    {
        // base cases
        if (high < low)
            return -1;
        if (high == low)
            return low;

        /* low + (high - low)/2; */
        int mid = (low + high) / 2;

        if (mid < high && arr[mid] > arr[mid + 1])
            return mid;

        if (mid > low && arr[mid] < arr[mid - 1])
            return (mid - 1);

        if (arr[low] >= arr[mid])
            return findPivot(arr, low, mid - 1);

        return findPivot(arr, mid + 1, high);
    }

    /* Standard Binary Search function */
    static int binarySearch(int[] arr, int low,
                            int high, int key)
    {
        if (high < low)
            return -1;

        /* low + (high - low)/2; */
        int mid = (low + high) / 2;

        if (key == arr[mid])
            return mid;
        if (key > arr[mid])
            return binarySearch(arr, (mid + 1), high, key);

        return binarySearch(arr, low, (mid - 1), key);
    }

    // Driver Code
    public static void Main()
    {
        // Let us search 3 in below array
        int[] arr1 = { 5, 6, 7, 8, 9, 10, 1, 2, 3 };
        int n = arr1.Length;
        int key = 3;
        Console.Write("Index of the element is : "
                      + pivotedBinarySearch(arr1, n, key));
    }
}

// This code is contributed by vt_m.

PHP

<?php
// PHP Program to search an element
// in a sorted and pivoted array

// Standard Binary Search function
function binarySearch($arr, $low,
                      $high, $key)
{
    if ($high < $low)
        return -1;

    /*low + (high - low)/2;*/
    $mid = floor($low + $high) / 2;

    if ($key == $arr[$mid])
        return $mid;

    if ($key > $arr[$mid])
        return binarySearch($arr, ($mid + 1),
                                $high, $key);

    else
        return binarySearch($arr, $low,
                      ($mid -1), $key);
}

// Function to get pivot.
// For array 3, 4, 5, 6, 1, 2
// it returns 3 (index of 6)
function findPivot($arr, $low, $high)
{

    // base cases
    if ($high < $low)
        return -1;
    if ($high == $low)
        return $low;

    /*low + (high - low)/2;*/
    $mid = ($low + $high)/2;
    if ($mid < $high and $arr[$mid] >
                     $arr[$mid + 1])
        return $mid;

    if ($mid > $low and $arr[$mid] <
                    $arr[$mid - 1])
        return ($mid - 1);

    if ($arr[$low] >= $arr[$mid])
        return findPivot($arr, $low,
                          $mid - 1);

    return findPivot($arr, $mid + 1, $high);
}

// Searches an element key
// in a pivoted sorted array
// arr[] of size n */
function pivotedBinarySearch($arr, $n, $key)
{
    $pivot = findPivot($arr, 0, $n - 1);

    // If we didn't find a pivot,
    // then array is not rotated
    // at all
    if ($pivot == -1)
        return binarySearch($arr, 0,
                       $n - 1, $key);

    // If we found a pivot,
    // then first compare
    // with pivot and then
    // search in two subarrays
    // around pivot
    if ($arr[$pivot] == $key)
        return $pivot;

    if ($arr[0] <= $key)
        return binarySearch($arr, 0,
                   $pivot - 1, $key);

        return binarySearch($arr, $pivot + 1,
                                $n - 1, $key);
}

// Driver Code
// Let us search 3
// in below array
$arr1 = array(5, 6, 7, 8, 9, 10, 1, 2, 3);
$n = count($arr1);
$key = 3;

// Function calling
echo "Index of the element is : ",
      pivotedBinarySearch($arr1, $n, $key);

// This code is contributed by anuj_67.
?>

Javascript

<script>
/* JavaScript Program to search an element
   in a sorted and pivoted array*/

/* Standard Binary Search function*/
function binarySearch( arr, low,
                  high, key){
    if (high < low)
        return -1;

    let mid = Math.floor((low + high) / 2); /*low + (high - low)/2;*/
    if (key == arr[mid])
        return mid;

    if (key > arr[mid])
        return binarySearch(arr, (mid + 1), high, key);

    // else
    return binarySearch(arr, low, (mid - 1), key);
}

/* Function to get pivot. For array 3, 4, 5, 6, 1, 2
   it returns 3 (index of 6) */
function findPivot( arr, low, high){
    // base cases
    if (high < low)
        return -1;
    if (high == low)
        return low;

    let mid = Math.floor((low + high) / 2); /*low + (high - low)/2;*/
    if (mid < high && arr[mid] > arr[mid + 1])
        return mid;

    if (mid > low && arr[mid] < arr[mid - 1])
        return (mid - 1);

    if (arr[low] >= arr[mid])
        return findPivot(arr, low, mid - 1);

    return findPivot(arr, mid + 1, high);
}

/* Searches an element key in a pivoted
   sorted array arr[] of size n */
function pivotedBinarySearch( arr, n, key){
    let pivot = findPivot(arr, 0, n - 1);

    // If we didn't find a pivot,
    // then array is not rotated at all
    if (pivot == -1)
        return binarySearch(arr, 0, n - 1, key);

    // If we found a pivot, then first compare with pivot
    // and then search in two subarrays around pivot
    if (arr[pivot] == key)
        return pivot;

    if (arr[0] <= key)
        return binarySearch(arr, 0, pivot - 1, key);

    return binarySearch(arr, pivot + 1, n - 1, key);
}

/* Driver program to check above functions */
// Let us search 3 in below array
let arr1 = [ 5, 6, 7, 8, 9, 10, 1, 2, 3 ];
let n = arr1.length;
let key = 3;
// Function calling
document.write( "Index of the element is : "
         + pivotedBinarySearch(arr1, n, key));

</script>

Output

Index of the element is : 8

Time Complexity: O(log N) Binary Search requires log n comparisons to find the element.
Auxiliary Complexity: O(1)

Thanks to Ajay Mishra for providing the above solution.

Approach 2 (Direct Binary Search on Array without finding Pivot):

The idea is to instead of two or more passes of binary search, the result can be found in one pass of binary search. 

The idea is to create a recursive function to implement the binary search where the search region is [l, r]. For each recursive call:

• We calculate the mid value as mid = (l + h) / 2
• Then try to figure out if l to mid is sorted, or (mid+1) to h is sorted
• Based on that decide the next search region and keep on doing this till the element is found or l overcomes h.

Follow the steps mentioned below to implement the idea:  

• Use a recursive function to implement binary search to find the key:
  • Find middle-point mid = (l + h)/2
  • If the key is present at the middle point, return mid.
  • Else if the value at l is less than the one at mid then arr[l . . . mid] is sorted
    • If the key to be searched lies in the range from arr[l] to arr[mid], recur for arr[l . . . mid].
    • Else recur for arr[mid+1 . . . h]  
  • Else arr[mid+1. . . h] is sorted: 
    • If the key to be searched lies in the range from arr[mid+1] to arr[h], recur for arr[mid+1. . . h].
    • Else recur for arr[l. . . mid] 

Follow the below illustration for a better understanding:

Illustration:

Input arr[] = {3, 4, 5, 1, 2}, key = 1
Initially low = 0, high = 4.

low = 0, high = 4:
        => mid = 2
        => arr[mid] = 5, which is not the desired value.
        => arr[low] < arr[mid] So, the left half is sorted.
        => key < arr[low], So the next search region is 3 to 4.

low = 3, high = 4:
        => mid = 3
        => arr[mid] = 1 = key
        => So the element is found at index 3.

The element is found at index 3.

Below is the implementation of the above idea:  

C++

// Search an element in sorted and rotated
// array using single pass of Binary Search
#include <bits/stdc++.h>
using namespace std;

// Returns index of key in arr[l..h] if
// key is present, otherwise returns -1
int search(int arr[], int l, int h, int key)
{
    if (l > h)
        return -1;

    int mid = (l + h) / 2;
    if (arr[mid] == key)
        return mid;

    /* If arr[l...mid] is sorted */
    if (arr[l] <= arr[mid]) {
        /* As this subarray is sorted, we can quickly
        check if key lies in half or other half */
        if (key >= arr[l] && key <= arr[mid])
            return search(arr, l, mid - 1, key);
        /*If key not lies in first half subarray,
           Divide other half  into two subarrays,
           such that we can quickly check if key lies
           in other half */
        return search(arr, mid + 1, h, key);
    }

    /* If arr[l..mid] first subarray is not sorted, then
    arr[mid... h] must be sorted subarray */
    if (key >= arr[mid] && key <= arr[h])
        return search(arr, mid + 1, h, key);

    return search(arr, l, mid - 1, key);
}

// Driver program
int main()
{
    int arr[] = { 4, 5, 6, 7, 8, 9, 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int key = 3;
    int i = search(arr, 0, n - 1, key);

    if (i != -1)
        cout << "Index: " << i << endl;
    else
        cout << "Key not found";
}

// This code is contributed by Aditya Kumar (adityakumar129)

C

// Search an element in sorted and rotated
// array using single pass of Binary Search
#include <stdio.h>

// Returns index of key in arr[l..h] if
// key is present, otherwise returns -1
int search(int arr[], int l, int h, int key)
{
    if (l > h)
        return -1;

    int mid = (l + h) / 2;
    if (arr[mid] == key)
        return mid;

    /* If arr[l...mid] is sorted */
    if (arr[l] <= arr[mid]) {
        /* As this subarray is sorted, we can quickly
        check if key lies in half or other half */
        if (key >= arr[l] && key <= arr[mid])
            return search(arr, l, mid - 1, key);
        /*If key not lies in first half subarray,
           Divide other half  into two subarrays,
           such that we can quickly check if key lies
           in other half */
        return search(arr, mid + 1, h, key);
    }

    /* If arr[l..mid] first subarray is not sorted, then
    arr[mid... h] must be sorted subarray */
    if (key >= arr[mid] && key <= arr[h])
        return search(arr, mid + 1, h, key);

    return search(arr, l, mid - 1, key);
}

// Driver program
int main()
{
    int arr[] = { 4, 5, 6, 7, 8, 9, 1, 2, 3 };
    int n = sizeof(arr) / sizeof(arr[0]);
    int key = 3;
    int i = search(arr, 0, n - 1, key);

    if (i != -1)
        printf("Index: %d\n", i);
    else
        printf("Key not found");
}

// This code is contributed by Aditya Kumar (adityakumar129)

Java

/* Java program to search an element in
   sorted and rotated array using
   single pass of Binary Search*/
import java.io.*;

class Main {
    // Returns index of key in arr[l..h]
    // if key is present, otherwise returns -1
    static int search(int arr[], int l, int h, int key)
    {
        if (l > h)
            return -1;

        int mid = (l + h) / 2;
        if (arr[mid] == key)
            return mid;

        /* If arr[l...mid] first subarray is sorted */
        if (arr[l] <= arr[mid]) {
            /* As this subarray is sorted, we
               can quickly check if key lies in
               half or other half */
            if (key >= arr[l] && key <= arr[mid])
                return search(arr, l, mid - 1, key);
            /*If key not lies in first half subarray,
           Divide other half  into two subarrays,
           such that we can quickly check if key lies
           in other half */
            return search(arr, mid + 1, h, key);
        }

        /* If arr[l..mid] first subarray is not sorted,
           then arr[mid... h] must be sorted subarray*/
        if (key >= arr[mid] && key <= arr[h])
            return search(arr, mid + 1, h, key);

        return search(arr, l, mid - 1, key);
    }

    // main function
    public static void main(String args[])
    {
        int arr[] = { 4, 5, 6, 7, 8, 9, 1, 2, 3 };
        int n = arr.length;
        int key = 3;
        int i = search(arr, 0, n - 1, key);
        if (i != -1)
            System.out.println("Index: " + i);
        else
            System.out.println("Key not found");
    }
}

// This code is contributed by Aditya Kumar (adityakumar129)

Python3

# Search an element in sorted and rotated array using
# single pass of Binary Search

# Returns index of key in arr[l..h] if key is present,
# otherwise returns -1
def search(arr, l, h, key):
    if l > h:
        return -1

    mid = (l + h) // 2
    if arr[mid] == key:
        return mid

    # If arr[l...mid] is sorted
    if arr[l] <= arr[mid]:

        # As this subarray is sorted, we can quickly
        # check if key lies in half or other half
        if key >= arr[l] and key <= arr[mid]:
            return search(arr, l, mid-1, key)
        return search(arr, mid + 1, h, key)

    # If arr[l..mid] is not sorted, then arr[mid... r]
    # must be sorted
    if key >= arr[mid] and key <= arr[h]:
        return search(arr, mid + 1, h, key)
    return search(arr, l, mid-1, key)


# Driver program
if __name__ == '__main__':
    arr = [4, 5, 6, 7, 8, 9, 1, 2, 3]
    key = 3
    i = search(arr, 0, len(arr)-1, key)
    if i != -1:
        print("Index: % d" % i)
    else:
        print("Key not found")

# This code is contributed by Shreyanshi Arun

C#

/* C# program to search an element in
sorted and rotated array using
single pass of Binary Search*/
using System;

class GFG {

    // Returns index of key in arr[l..h]
    // if key is present, otherwise
    // returns -1
    static int search(int[] arr, int l, int h, int key)
    {
        if (l > h)
            return -1;

        int mid = (l + h) / 2;

        if (arr[mid] == key)
            return mid;

        /* If arr[l...mid] is sorted */
        if (arr[l] <= arr[mid]) {

            /* As this subarray is sorted, we
            can quickly check if key lies in
            half or other half */
            if (key >= arr[l] && key <= arr[mid])
                return search(arr, l, mid - 1, key);

            return search(arr, mid + 1, h, key);
        }

        /* If arr[l..mid] is not sorted,
        then arr[mid... r] must be sorted*/
        if (key >= arr[mid] && key <= arr[h])
            return search(arr, mid + 1, h, key);

        return search(arr, l, mid - 1, key);
    }

    // main function
    public static void Main()
    {
        int[] arr = { 4, 5, 6, 7, 8, 9, 1, 2, 3 };
        int n = arr.Length;
        int key = 3;
        int i = search(arr, 0, n - 1, key);

        if (i != -1)
            Console.WriteLine("Index: " + i);
        else
            Console.WriteLine("Key not found");
    }
}

// This code is contributed by anuj_67.

PHP

<?php
// Search an element in sorted and rotated
// array using single pass of Binary Search

// Returns index of key in arr[l..h] if
// key is present, otherwise returns -1
function search($arr, $l, $h, $key)
{
    if ($l > $h) return -1;

    $mid = floor(($l + $h) / 2);
    if ($arr[$mid] == $key)
        return $mid;

    /* If arr[l...mid] is sorted */
    if ($arr[$l] <= $arr[$mid])
    {

        /* As this subarray is
           sorted, we can quickly
           check if key lies in
           half or other half */
        if ($key >= $arr[$l] and
            $key <= $arr[$mid])
                return search($arr, $l,
                       $mid - 1, $key);

        return search($arr, $mid + 1,
                           $h, $key);
    }

    /* If arr[l..mid] is not
       sorted, then arr[mid... r]
       must be sorted*/
    if ($key >= $arr[$mid] and
          $key <= $arr[$h])
        return search($arr, $mid + 1,
                            $h, $key);

    return search($arr, $l,
             $mid-1, $key);
}

    // Driver Code
    $arr = array( 5, 6, 7, 8, 9, 10, 1, 2, 3 );
    $n = sizeof($arr);
    $key = 3;
    $i = search($arr, 0, $n-1, $key);

    if ($i != -1)
        echo "Index: ", $i, " \n";
    else
        echo "Key not found";

// This code is contributed by ajit
?>

Javascript

<script>

// Search an element in sorted and rotated
// array using single pass of Binary Search

// Returns index of key in arr[l..h] if
// key is present, otherwise returns -1
function search(arr, l, h, key){
    if (l > h)
        return -1;

    let mid = Math.floor((l + h) / 2);
    if (arr[mid] == key)
        return mid;

    /* If arr[l...mid] is sorted */
    if (arr[l] <= arr[mid]) {
        /* As this subarray is sorted, we can quickly
        check if key lies in half or other half */
        if (key >= arr[l] && key <= arr[mid])
            return search(arr, l, mid - 1, key);
        /*If key not lies in first half subarray,
           Divide other half  into two subarrays,
           such that we can quickly check if key lies
           in other half */
        return search(arr, mid + 1, h, key);
    }

    /* If arr[l..mid] first subarray is not sorted,
    then arr[mid... h]
    must be sorted subarray */
    if (key >= arr[mid] && key <= arr[h])
        return search(arr, mid + 1, h, key);

    return search(arr, l, mid - 1, key);
}

// Driver program
let arr = [ 4, 5, 6, 7, 8, 9, 1, 2, 3 ];
let n = arr.length;
let key = 3;
let i = search(arr, 0, n - 1, key);
if (i != -1)
    document.write("Index: " +i +"\n");
else
    document.write("Key not found");

</script>

Output

Index: 8

Time Complexity: O(log N). Binary Search requires log n comparisons to find the element. So time complexity is O(log n).
Auxiliary Space: O(1). As no extra space is required.

Thanks to Gaurav Ahirwar for suggesting the above solution.

How to handle duplicates?

At first look, it doesn’t look possible to search in O(Log N) time in all cases when duplicates are allowed. 
For example consider searching 0 in {2, 2, 2, 2, 2, 2, 2, 2, 0, 2} and {2, 0, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2}. 
Look into the following article to find a solution to this issue: https://www.geeksforgeeks.org/search-an-element-in-a-sorted-and-rotated-array-with-duplicates/

Similar Articles:

• Find the minimum element in a sorted and rotated array
• Given a sorted and rotated array, find if there is a pair with a given sum.

Please write comments if you find any bug in the above codes/algorithms, or find other ways to solve the same problem.

Approach#3: Using linear search

This approach uses linear search to find the index of the key in a sorted and rotated array. The idea is to iterate through the array and compare each element with the key until we find a match.

Algorithm

1. Initialize a loop index i to 0, and iterate over the array using a for loop.
2. For each element of the array, check if it is equal to the given key.
3. If the key is found, return the index of the element.
4. If the end of the array is reached without finding the key, return -1 to indicate that the key was not found.

Python3

def search_rotated_array(arr, key):
    n = len(arr)
    for i in range(n):
        if arr[i] == key:
            return i
    return -1

arr = [5, 6, 7, 8, 9, 10, 1, 2, 3]
key = 3
index = search_rotated_array(arr, key)
if index != -1:
    print(f"Found at index {index}")
else:
    print("Not found")

Output

Found at index 8

Time complexity of this algorithm is O(n), where n is the length of the input array. 

Space complexity is O(1), as the program only uses a constant amount of extra space to store the loop index and the return value.