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Code Recursion

Find an element in a sorted and rotated array

In computer science, the problem of finding an element in a sorted and rotated array is a common challenge. The array is initially sorted in ascending order and then rotated at some pivot point, where the elements to the right of the pivot come before the elements to the left. The task is to efficiently locate a given target element in the rotated array.

Problem Statement

Given a sorted and rotated array arr, we need to find the index of a target element k in the array. If the element is present, the algorithm should return the index of the target element; otherwise, it should return -1.

Explanation with Example

Let's take the array arr = [7, 8, 9, 1, 2, 3, 4, 5, 6] as an example. The array is sorted in ascending order initially, but it has been rotated at the pivot point between 9 and 1. If we want to find the element 2, the algorithm should return index 4. Similarly, for element 9, the algorithm should return index 2.

Pseudocode

The following is the standard pseudocode for the algorithm to find an element in a sorted and rotated array:

1. Function binarySearch(arr, k, low, high)
2.     If low > high
3.         Return -1
4.     Else
5.         mid = (low + high) / 2
6.         If arr[mid] == k
7.             Return mid
8.         If arr[low] < arr[mid]
9.             If k >= arr[low] AND k < arr[mid]
10.                Return binarySearch(arr, k, low, mid - 1)
11.            Else
12.                Return binarySearch(arr, k, mid + 1, high)
13.        Else If arr[mid] < arr[high]
14.            If k > arr[mid] AND k <= arr[high]
15.                Return binarySearch(arr, k, mid + 1, high)
16.            Else
17.                Return binarySearch(arr, k, low, mid - 1)
18.        Else If arr[low] == arr[mid]
19.            If arr[mid] != arr[high]
20.                Return binarySearch(arr, k, mid + 1, high)
21.            Else
22.                result = binarySearch(arr, k, low, mid - 1)
23.                If result == -1
24.                    result = binarySearch(arr, k, mid + 1, high)
25.                Return result
    

Algorithm Explanation

The algorithm is a variation of the binary search algorithm. It takes a sorted and rotated array arr, the target element k, and the low and high indices of the array as input. It searches for the target element k in the array and returns the index of the element if found, or -1 if not found.

The binarySearch function first checks if the low index is greater than the high index, which means the target element is not present, and it returns -1. Otherwise, it calculates the mid index and compares the element at the mid index with the target element k.

If the element at the mid index is equal to the target element k, it means the element is found, and the function returns the mid index.

If the element at the low index is less than the element at the mid index, it implies that the elements from low to mid are in sorted order. In this case, the algorithm checks if the target element k falls within this range. If it does, the algorithm performs a recursive binary search on the left half of the array (from low to mid-1); otherwise, it performs a recursive binary search on the right half of the array (from mid+1 to high).

Similarly, if the element at the mid index is less than the element at the high index, it means the elements from mid to high are in sorted order. The algorithm checks if the target element k falls within this range and proceeds with the recursive binary search accordingly.

If the elements at both the low and mid indices are equal, it indicates that there are repeated elements in the array. The algorithm checks if the element at the mid index is different from the element at the high index to decide which side to perform the recursive binary search. If the element at the mid index is also equal to the element at the high index, it means we need to check both the left and right subtrees to find the target element.

Program Solution

// C Program
// Find an element in a sorted and rotated array
#include <stdio.h>

// Display array elements
void display(int arr[], int size)
{
	printf("\n\n Array Elements \n [");
	for (int i = 0; i < size; ++i)
	{
		printf("   %d", arr[i]);
	}
	printf("  ]\n");
}
// Finding the element location of sorted rotated array
int binarySearch(int arr[], int k, int low, int high)
{
	if (low > high)
	{
		return -1;
	}
	else
	{
		int mid = (low + high) / 2;
		if (arr[mid] == k)
		{
			// When element exist
			return mid;
		}
		if (arr[low] < arr[mid])
		{
			// When elements are sort in location low to mid position
			if (k >= arr[low] && k < arr[mid])
			{
				// When the element is possible in the (low) to (mid-1) position
				return binarySearch(arr, k, low, mid - 1);
			}
			else
			{
				// When the element is possible in the (mid+1) to (high) position
				return binarySearch(arr, k, mid + 1, high);
			}
		}
		else if (arr[mid] < arr[high])
		{
			if (k > arr[mid] && k <= arr[high])
			{
				// When the element is possible in the (mid+1) to (high) position
				return binarySearch(arr, k, mid + 1, high);
			}
			else
			{
				// When the element is possible in the (low) to (mid-1) position
				return binarySearch(arr, k, low, mid - 1);
			}
		}
		else if (arr[low] == arr[mid])
		{
			// When repeated array element exists
			if (arr[mid] != arr[high])
			{
				return binarySearch(arr, k, mid + 1, high);
			}
			else
			{
				// Check if that element exists in left subtree
				int result = binarySearch(arr, k, low, mid - 1);
				if (result == -1)
				{
					// When element not exist in left subtree then 
					// Check that element exists in right subtree
					result = binarySearch(arr, k, mid + 1, high);
				}
				return result;
			}
		}
	}
}
void findElement(int arr[], int size, int k)
{
	int location = binarySearch(arr, k, 0, size - 1);
	if (location == -1)
	{
		printf("\n Element %d Are not exist ", k);
	}
	else
	{
		printf("\n Element %d exist in location %d ", k, location);
	}
}
int main()
{
	// Defining sorted and rotated array of integer element
	int arr1[] = {
		7 , 8 , 9 , 1 , 2 , 3 , 4 , 5 , 6
	};
	int arr2[] = {
		11 , 22 , 43 , 45 , 51 , 62 , 73 , 1 , 9 , 10
	};
	// Get the size
	int size = sizeof(arr1) / sizeof(arr1[0]);
	display(arr1, size);
	findElement(arr1, size, 9);
	findElement(arr1, size, 3);
	// Get the size
	size = sizeof(arr2) / sizeof(arr2[0]);
	display(arr2, size);
	// Test Case B (of arr2)
	findElement(arr2, size, 9);
	findElement(arr2, size, 11);
	findElement(arr2, size, 67);
	return 0;
}

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6  ]

 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10  ]

 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist
/*
    Java program 
    Find an element in a sorted and rotated array
*/
public class SearchElement
{
    // Display array elements
    public void display(int[] arr, int size)
    {
        System.out.print("\n\n Array Elements \n [");
        for (int i = 0; i < size; ++i)
        {
            System.out.print("   " + arr[i]);
        }
        System.out.print(" ]\n");
    }
    // Finding the element location of sorted rotated array
    public int binarySearch(int[] arr, int k, int low, int high)
    {
        if (low > high)
        {
            return -1;
        }
        else
        {
            int mid = (low + high) / 2;
            if (arr[mid] == k)
            {
                // When element exist
                return mid;
            }
            if (arr[low] < arr[mid])
            {
                // When elements are sort in location low to mid position
                if (k >= arr[low] && k < arr[mid])
                {
                    // When the element is possible in the (low) to (mid-1) position
                    return binarySearch(arr, k, low, mid - 1);
                }
                else
                {
                    // When the element is possible in the (mid+1) to (high) position
                    return binarySearch(arr, k, mid + 1, high);
                }
            }
            else if (arr[mid] < arr[high])
            {
                if (k > arr[mid] && k <= arr[high])
                {
                    // When the element is possible in the (mid+1) to (high) position
                    return binarySearch(arr, k, mid + 1, high);
                }
                else
                {
                    // When the element is possible in the (low) to (mid-1) position
                    return binarySearch(arr, k, low, mid - 1);
                }
            }
            else if (arr[low] == arr[mid])
            {
                // When repeated array element exists
                if (arr[mid] != arr[high])
                {
                    return binarySearch(arr, k, mid + 1, high);
                }
                else
                {
                    // Check if that element exists in left subtree
                    int result = binarySearch(arr, k, low, mid - 1);
                    if (result == -1)
                    {
                        // When element not exist in left subtree then 
                        // Check that element exists in right subtree
                        result = binarySearch(arr, k, mid + 1, high);
                    }
                    return result;
                }
            }
        }
        return -1;
    }
    // Handle request to find array elements
    public void findElement(int[] arr, int size, int k)
    {
        int location = binarySearch(arr, k, 0, size - 1);
        if (location == -1)
        {
            System.out.print("\n Element " + k + " Are not exist ");
        }
        else
        {
            System.out.print("\n Element " + k + " exist in location " + location);
        }
    }
    public static void main(String[] args)
    {
        SearchElement task = new SearchElement();
        // Defining sorted and rotated array of integer element
        int[] arr1 = {
            7 , 8 , 9 , 1 , 2 , 3 , 4 , 5 , 6
        };
        int[] arr2 = {
            11 , 22 , 43 , 45 , 51 , 62 , 73 , 1 , 9 , 10
        };
        // Get the size
        int size = arr1.length;
        task.display(arr1, size);
        // Test Cases
        task.findElement(arr1, size, 9);
        task.findElement(arr1, size, 3);
        // Get the size
        size = arr2.length;
        task.display(arr2, size);
        // Test Cases
        task.findElement(arr2, size, 9);
        task.findElement(arr2, size, 11);
        task.findElement(arr2, size, 67);
    }
}

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6 ]

 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10 ]

 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist
// Include header file
#include <iostream>

using namespace std;
/*
    C++ program 
    Find an element in a sorted and rotated array
*/
class SearchElement
{
    public:
    //  Display array elements
    void display(int arr[], int size)
    {
        cout << "\n\n Array Elements \n [";
        for (int i = 0; i < size; ++i)
        {
            cout << "   " << arr[i];
        }
        cout << " ]";
    }
    //  Finding the element location of sorted rotated array
    int binarySearch(int arr[], int k, int low, int high)
    {
        if (low > high)
        {
            return -1;
        }
        else
        {
            int mid = (low + high) / 2;
            if (arr[mid] == k)
            {
                //  When element exist
                return mid;
            }
            if (arr[low] < arr[mid])
            {
                //  When elements are sort in location low to mid position
                if (k >= arr[low] && k < arr[mid])
                {
                    //  When the element is possible in the (low) to (mid-1) position
                    return this->binarySearch(arr, k, low, mid - 1);
                }
                else
                {
                    //  When the element is possible in the (mid+1) to (high) position
                    return this->binarySearch(arr, k, mid + 1, high);
                }
            }
            else if (arr[mid] < arr[high])
            {
                if (k > arr[mid] && k <= arr[high])
                {
                    //  When the element is possible in the (mid+1) to (high) position
                    return this->binarySearch(arr, k, mid + 1, high);
                }
                else
                {
                    //  When the element is possible in the (low) to (mid-1) position
                    return this->binarySearch(arr, k, low, mid - 1);
                }
            }
            else if (arr[low] == arr[mid])
            {
                //  When repeated array element exists
                if (arr[mid] != arr[high])
                {
                    return this->binarySearch(arr, k, mid + 1, high);
                }
                else
                {
                    //  Check if that element exists in left subtree
                    int result = this->binarySearch(arr, k, low, mid - 1);
                    if (result == -1)
                    {
                        //  When element not exist in left subtree then
                        //  Check that element exists in right subtree
                        result = this->binarySearch(arr, k, mid + 1, high);
                    }
                    return result;
                }
            }
        }
        return -1;
    }
    //  Handle request to find array elements
    void findElement(int arr[], int size, int k)
    {
        int location = this->binarySearch(arr, k, 0, size - 1);
        if (location == -1)
        {
            cout << "\n Element " << k << " Are not exist ";
        }
        else
        {
            cout << "\n Element " << k << " exist in location " << location;
        }
    }
};
int main()
{
    SearchElement task = SearchElement();
    //  Defining sorted and rotated array of integer element
    int arr1[] = {
        7 , 8 , 9 , 1 , 2 , 3 , 4 , 5 , 6
    };
    int arr2[] = {
        11 , 22 , 43 , 45 , 51 , 62 , 73 , 1 , 9 , 10
    };
    //  Get the size
    int size = sizeof(arr1) / sizeof(arr1[0]);
    task.display(arr1, size);
    //  Test Cases
    task.findElement(arr1, size, 9);
    task.findElement(arr1, size, 3);
    //  Get the size
    size = sizeof(arr2) / sizeof(arr2[0]);
    task.display(arr2, size);
    //  Test Cases
    task.findElement(arr2, size, 9);
    task.findElement(arr2, size, 11);
    task.findElement(arr2, size, 67);
    return 0;
}

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6 ]
 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10 ]
 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist
// Include namespace system
using System;
/*
    C# program 
    Find an element in a sorted and rotated array
*/
public class SearchElement
{
	//  Display array elements
	public void display(int[] arr, int size)
	{
		Console.Write("\n\n Array Elements \n [");
		for (int i = 0; i < size; ++i)
		{
			Console.Write("   " + arr[i]);
		}
		Console.Write(" ]");
	}
	//  Finding the element location of sorted rotated array
	public int binarySearch(int[] arr, int k, int low, int high)
	{
		if (low > high)
		{
			return -1;
		}
		else
		{
			int mid = (low + high) / 2;
			if (arr[mid] == k)
			{
				//  When element exist
				return mid;
			}
			if (arr[low] < arr[mid])
			{
				//  When elements are sort in location low to mid position
				if (k >= arr[low] && k < arr[mid])
				{
					//  When the element is possible in the (low) to (mid-1) position
					return binarySearch(arr, k, low, mid - 1);
				}
				else
				{
					//  When the element is possible in the (mid+1) to (high) position
					return binarySearch(arr, k, mid + 1, high);
				}
			}
			else if (arr[mid] < arr[high])
			{
				if (k > arr[mid] && k <= arr[high])
				{
					//  When the element is possible in the (mid+1) to (high) position
					return binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					//  When the element is possible in the (low) to (mid-1) position
					return binarySearch(arr, k, low, mid - 1);
				}
			}
			else if (arr[low] == arr[mid])
			{
				//  When repeated array element exists
				if (arr[mid] != arr[high])
				{
					return binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					//  Check if that element exists in left subtree
					int result = binarySearch(arr, k, low, mid - 1);
					if (result == -1)
					{
						//  When element not exist in left subtree then
						//  Check that element exists in right subtree
						result = binarySearch(arr, k, mid + 1, high);
					}
					return result;
				}
			}
		}
		return -1;
	}
	//  Handle request to find array elements
	public void findElement(int[] arr, int size, int k)
	{
		int location = binarySearch(arr, k, 0, size - 1);
		if (location == -1)
		{
			Console.Write("\n Element " + k + " Are not exist ");
		}
		else
		{
			Console.Write("\n Element " + k + " exist in location " + location);
		}
	}
	public static void Main(String[] args)
	{
		SearchElement task = new SearchElement();
		//  Defining sorted and rotated array of integer element
		int[] arr1 = {
			7 , 8 , 9 , 1 , 2 , 3 , 4 , 5 , 6
		};
		int[] arr2 = {
			11 , 22 , 43 , 45 , 51 , 62 , 73 , 1 , 9 , 10
		};
		//  Get the size
		int size = arr1.Length;
		task.display(arr1, size);
		//  Test Cases
		task.findElement(arr1, size, 9);
		task.findElement(arr1, size, 3);
		//  Get the size
		size = arr2.Length;
		task.display(arr2, size);
		//  Test Cases
		task.findElement(arr2, size, 9);
		task.findElement(arr2, size, 11);
		task.findElement(arr2, size, 67);
	}
}

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6 ]
 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10 ]
 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist
<?php
/*
    Php program 
    Find an element in a sorted and rotated array
*/
class SearchElement
{
	//  Display array elements
	public	function display( & $arr, $size)
	{
		echo "\n\n Array Elements \n [";
		for ($i = 0; $i < $size; ++$i)
		{
			echo "   ". $arr[$i];
		}
		echo " ]";
	}
	//  Finding the element location of sorted rotated array
	public	function binarySearch( & $arr, $k, $low, $high)
	{
		if ($low > $high)
		{
			return -1;
		}
		else
		{
			$mid = intval(($low + $high) / 2);
			if ($arr[$mid] == $k)
			{
				//  When element exist
				return $mid;
			}
			if ($arr[$low] < $arr[$mid])
			{
				//  When elements are sort in location low to mid position
				if ($k >= $arr[$low] && $k < $arr[$mid])
				{
					//  When the element is possible in the (low) to (mid-1) position
					return $this->binarySearch($arr, $k, $low, $mid - 1);
				}
				else
				{
					//  When the element is possible in the (mid+1) to (high) position
					return $this->binarySearch($arr, $k, $mid + 1, $high);
				}
			}
			else if ($arr[$mid] < $arr[$high])
			{
				if ($k > $arr[$mid] && $k <= $arr[$high])
				{
					//  When the element is possible in the (mid+1) to (high) position
					return $this->binarySearch($arr, $k, $mid + 1, $high);
				}
				else
				{
					//  When the element is possible in the (low) to (mid-1) position
					return $this->binarySearch($arr, $k, $low, $mid - 1);
				}
			}
			else if ($arr[$low] == $arr[$mid])
			{
				//  When repeated array element exists
				if ($arr[$mid] != $arr[$high])
				{
					return $this->binarySearch($arr, $k, $mid + 1, $high);
				}
				else
				{
					//  Check if that element exists in left subtree
					$result = $this->binarySearch($arr, $k, $low, $mid - 1);
					if ($result == -1)
					{
						//  When element not exist in left subtree then
						//  Check that element exists in right subtree
						$result = $this->binarySearch($arr, $k, $mid + 1, $high);
					}
					return $result;
				}
			}
		}
		return -1;
	}
	//  Handle request to find array elements
	public	function findElement( & $arr, $size, $k)
	{
		$location = $this->binarySearch($arr, $k, 0, $size - 1);
		if ($location == -1)
		{
			echo "\n Element ". $k ." Are not exist ";
		}
		else
		{
			echo "\n Element ". $k ." exist in location ". $location;
		}
	}
}

function main()
{
	$task = new SearchElement();
	//  Defining sorted and rotated array of integer element
	$arr1 = array(7, 8, 9, 1, 2, 3, 4, 5, 6);
	$arr2 = array(11, 22, 43, 45, 51, 62, 73, 1, 9, 10);
	//  Get the size
	$size = count($arr1);
	$task->display($arr1, $size);
	//  Test Cases
	$task->findElement($arr1, $size, 9);
	$task->findElement($arr1, $size, 3);
	//  Get the size
	$size = count($arr2);
	$task->display($arr2, $size);
	//  Test Cases
	$task->findElement($arr2, $size, 9);
	$task->findElement($arr2, $size, 11);
	$task->findElement($arr2, $size, 67);
}
main();

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6 ]
 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10 ]
 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist
/*
    Node Js program 
    Find an element in a sorted and rotated array
*/
class SearchElement
{
	//  Display array elements
	display(arr, size)
	{
		process.stdout.write("\n\n Array Elements \n [");
		for (var i = 0; i < size; ++i)
		{
			process.stdout.write("   " + arr[i]);
		}
		process.stdout.write(" ]");
	}
	//  Finding the element location of sorted rotated array
	binarySearch(arr, k, low, high)
	{
		if (low > high)
		{
			return -1;
		}
		else
		{
			var mid = parseInt((low + high) / 2);
			if (arr[mid] == k)
			{
				//  When element exist
				return mid;
			}
			if (arr[low] < arr[mid])
			{
				//  When elements are sort in location low to mid position
				if (k >= arr[low] && k < arr[mid])
				{
					//  When the element is possible in the (low) to (mid-1) position
					return this.binarySearch(arr, k, low, mid - 1);
				}
				else
				{
					//  When the element is possible in the (mid+1) to (high) position
					return this.binarySearch(arr, k, mid + 1, high);
				}
			}
			else if (arr[mid] < arr[high])
			{
				if (k > arr[mid] && k <= arr[high])
				{
					//  When the element is possible in the (mid+1) to (high) position
					return this.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					//  When the element is possible in the (low) to (mid-1) position
					return this.binarySearch(arr, k, low, mid - 1);
				}
			}
			else if (arr[low] == arr[mid])
			{
				//  When repeated array element exists
				if (arr[mid] != arr[high])
				{
					return this.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					//  Check if that element exists in left subtree
					var result = this.binarySearch(arr, k, low, mid - 1);
					if (result == -1)
					{
						//  When element not exist in left subtree then
						//  Check that element exists in right subtree
						result = this.binarySearch(arr, k, mid + 1, high);
					}
					return result;
				}
			}
		}
		return -1;
	}
	//  Handle request to find array elements
	findElement(arr, size, k)
	{
		var location = this.binarySearch(arr, k, 0, size - 1);
		if (location == -1)
		{
			process.stdout.write("\n Element " + k + " Are not exist ");
		}
		else
		{
			process.stdout.write("\n Element " + k + " exist in location " + location);
		}
	}
}

function main()
{
	var task = new SearchElement();
	//  Defining sorted and rotated array of integer element
	var arr1 = [7, 8, 9, 1, 2, 3, 4, 5, 6];
	var arr2 = [11, 22, 43, 45, 51, 62, 73, 1, 9, 10];
	//  Get the size
	var size = arr1.length;
	task.display(arr1, size);
	//  Test Cases
	task.findElement(arr1, size, 9);
	task.findElement(arr1, size, 3);
	//  Get the size
	size = arr2.length;
	task.display(arr2, size);
	//  Test Cases
	task.findElement(arr2, size, 9);
	task.findElement(arr2, size, 11);
	task.findElement(arr2, size, 67);
}
main();

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6 ]
 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10 ]
 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist
#   Python 3 program 
#   Find an element in a sorted and rotated list

class SearchElement :
	#   Display array elements
	def display(self, arr, size) :
		print("\n\n Array Elements \n [", end = "")
		i = 0
		while (i < size) :
			print("   ", arr[i], end = "")
			i += 1
		
		print(" ]", end = "")
	
	#   Finding the element location of sorted rotated array
	def binarySearch(self, arr, k, low, high) :
		if (low > high) :
			return -1
		else :
			mid = int((low + high) / 2)
			if (arr[mid] == k) :
				#   When element exist
				return mid
			
			if (arr[low] < arr[mid]) :
				#   When elements are sort in location low to mid position
				if (k >= arr[low] and k < arr[mid]) :
					#   When the element is possible in the (low) to (mid-1) position
					return self.binarySearch(arr, k, low, mid - 1)
				else :
					#   When the element is possible in the (mid+1) to (high) position
					return self.binarySearch(arr, k, mid + 1, high)
				
			
			elif(arr[mid] < arr[high]) :
				if (k > arr[mid] and k <= arr[high]) :
					#   When the element is possible in the (mid+1) to (high) position
					return self.binarySearch(arr, k, mid + 1, high)
				else :
					#   When the element is possible in the (low) to (mid-1) position
					return self.binarySearch(arr, k, low, mid - 1)
				
			
			elif(arr[low] == arr[mid]) :
				#   When repeated array element exists
				if (arr[mid] != arr[high]) :
					return self.binarySearch(arr, k, mid + 1, high)
				else :
					#   Check if that element exists in left subtree
					result = self.binarySearch(arr, k, low, mid - 1)
					if (result == -1) :
						#   When element not exist in left subtree then
						#   Check that element exists in right subtree
						result = self.binarySearch(arr, k, mid + 1, high)
					
					return result
				
			
		
		return -1
	
	#   Handle request to find array elements
	def findElement(self, arr, size, k) :
		location = self.binarySearch(arr, k, 0, size - 1)
		if (location == -1) :
			print("\n Element ", k ," Are not exist ", end = "")
		else :
			print("\n Element ", k ," exist in location ", location, end = "")
		
	

def main() :
	task = SearchElement()
	#   Defining sorted and rotated array of integer element
	arr1 = [7, 8, 9, 1, 2, 3, 4, 5, 6]
	arr2 = [11, 22, 43, 45, 51, 62, 73, 1, 9, 10]
	#   Get the size
	size = len(arr1)
	task.display(arr1, size)
	#   Test Cases
	task.findElement(arr1, size, 9)
	task.findElement(arr1, size, 3)
	#   Get the size
	size = len(arr2)
	task.display(arr2, size)
	#   Test Cases
	task.findElement(arr2, size, 9)
	task.findElement(arr2, size, 11)
	task.findElement(arr2, size, 67)

if __name__ == "__main__": main()

Output

 Array Elements
 [    7    8    9    1    2    3    4    5    6 ]
 Element  9  exist in location  2
 Element  3  exist in location  5

 Array Elements
 [    11    22    43    45    51    62    73    1    9    10 ]
 Element  9  exist in location  8
 Element  11  exist in location  0
 Element  67  Are not exist
#  Ruby program 
#  Find an element in a sorted and rotated array

class SearchElement 
	#   Display array elements
	def display(arr, size) 
		print("\n\n Array Elements \n [")
		i = 0
		while (i < size) 
			print("   ", arr[i])
			i += 1
		end

		print(" ]")
	end

	#   Finding the element location of sorted rotated array
	def binarySearch(arr, k, low, high) 
		if (low > high) 
			return -1
		else 
			mid = (low + high) / 2
			if (arr[mid] == k) 
				#   When element exist
				return mid
			end

			if (arr[low] < arr[mid]) 
				#   When elements are sort in location low to mid position
				if (k >= arr[low] && k < arr[mid]) 
					#   When the element is possible in the (low) to (mid-1) position
					return self.binarySearch(arr, k, low, mid - 1)
				else 
					#   When the element is possible in the (mid+1) to (high) position
					return self.binarySearch(arr, k, mid + 1, high)
				end

			elsif(arr[mid] < arr[high]) 
				if (k > arr[mid] && k <= arr[high]) 
					#   When the element is possible in the (mid+1) to (high) position
					return self.binarySearch(arr, k, mid + 1, high)
				else 
					#   When the element is possible in the (low) to (mid-1) position
					return self.binarySearch(arr, k, low, mid - 1)
				end

			elsif(arr[low] == arr[mid]) 
				#   When repeated array element exists
				if (arr[mid] != arr[high]) 
					return self.binarySearch(arr, k, mid + 1, high)
				else 
					#   Check if that element exists in left subtree
					result = self.binarySearch(arr, k, low, mid - 1)
					if (result == -1) 
						#   When element not exist in left subtree then
						#   Check that element exists in right subtree
						result = self.binarySearch(arr, k, mid + 1, high)
					end

					return result
				end

			end

		end

		return -1
	end

	#   Handle request to find array elements
	def findElement(arr, size, k) 
		location = self.binarySearch(arr, k, 0, size - 1)
		if (location == -1) 
			print("\n Element ", k ," Are not exist ")
		else 
			print("\n Element ", k ," exist in location ", location)
		end

	end

end

def main() 
	task = SearchElement.new()
	#   Defining sorted and rotated array of integer element
	arr1 = [7, 8, 9, 1, 2, 3, 4, 5, 6]
	arr2 = [11, 22, 43, 45, 51, 62, 73, 1, 9, 10]
	#   Get the size
	size = arr1.length
	task.display(arr1, size)
	#   Test Cases
	task.findElement(arr1, size, 9)
	task.findElement(arr1, size, 3)
	#   Get the size
	size = arr2.length
	task.display(arr2, size)
	#   Test Cases
	task.findElement(arr2, size, 9)
	task.findElement(arr2, size, 11)
	task.findElement(arr2, size, 67)
end

main()

Output


 Array Elements 
 [   7   8   9   1   2   3   4   5   6 ]
 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements 
 [   11   22   43   45   51   62   73   1   9   10 ]
 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist 
/*
    Scala program 
    Find an element in a sorted and rotated array
*/
class SearchElement
{
	//   Display array elements
	def display(arr: Array[Int], size: Int): Unit = {
		print("\n\n Array Elements \n [");
		var i: Int = 0;
		while (i < size)
		{
			print("   " + arr(i));
			i += 1;
		}
		print(" ]");
	}
	//   Finding the element location of sorted rotated array
	def binarySearch(arr: Array[Int], k: Int, low: Int, high: Int): Int = {
		if (low > high)
		{
			return -1;
		}
		else
		{
			var mid: Int = ((low + high) / 2).toInt;
			if (arr(mid) == k)
			{
				//   When element exist
				return mid;
			}
			if (arr(low) < arr(mid))
			{
				//   When elements are sort in location low to mid position
				if (k >= arr(low) && k < arr(mid))
				{
					//   When the element is possible in the (low) to (mid-1) position
					return this.binarySearch(arr, k, low, mid - 1);
				}
				else
				{
					//   When the element is possible in the (mid+1) to (high) position
					return this.binarySearch(arr, k, mid + 1, high);
				}
			}
			else if (arr(mid) < arr(high))
			{
				if (k > arr(mid) && k <= arr(high))
				{
					//   When the element is possible in the (mid+1) to (high) position
					return this.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					//   When the element is possible in the (low) to (mid-1) position
					return this.binarySearch(arr, k, low, mid - 1);
				}
			}
			else if (arr(low) == arr(mid))
			{
				//   When repeated array element exists
				if (arr(mid) != arr(high))
				{
					return this.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					//   Check if that element exists in left subtree
					var result: Int = this.binarySearch(arr, k, low, mid - 1);
					if (result == -1)
					{
						//   When element not exist in left subtree then
						//   Check that element exists in right subtree
						result = this.binarySearch(arr, k, mid + 1, high);
					}
					return result;
				}
			}
		}
		return -1;
	}
	//   Handle request to find array elements
	def findElement(arr: Array[Int], size: Int, k: Int): Unit = {
		var location: Int = this.binarySearch(arr, k, 0, size - 1);
		if (location == -1)
		{
			print("\n Element " + k + " Are not exist ");
		}
		else
		{
			print("\n Element " + k + " exist in location " + location);
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: SearchElement = new SearchElement();
		//   Defining sorted and rotated array of integer element
		var arr1: Array[Int] = Array(7, 8, 9, 1, 2, 3, 4, 5, 6);
		var arr2: Array[Int] = Array(11, 22, 43, 45, 51, 62, 73, 1, 9, 10);
		//   Get the size
		var size: Int = arr1.length;
		task.display(arr1, size);
		//   Test Cases
		task.findElement(arr1, size, 9);
		task.findElement(arr1, size, 3);
		//   Get the size
		size = arr2.length;
		task.display(arr2, size);
		//   Test Cases
		task.findElement(arr2, size, 9);
		task.findElement(arr2, size, 11);
		task.findElement(arr2, size, 67);
	}
}

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6 ]
 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10 ]
 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist
/*
    Swift 4 program 
    Find an element in a sorted and rotated array
*/
class SearchElement
{
	// Display array elements
	func display(_ arr: [Int], _ size: Int)
	{
		print("\n\n Array Elements \n [", terminator: "");
		var i: Int = 0;
		while (i < size)
		{
			print("   ", arr[i], terminator: "");
			i += 1;
		}
		print(" ]", terminator: "");
	}
	// Finding the element location of sorted rotated array
	func binarySearch(_ arr: [Int], _ k: Int, _ low: Int, _ high: Int)->Int
	{
		if (low > high)
		{
			return -1;
		}
		else
		{
			let mid: Int = (low + high) / 2;
			if (arr[mid] == k)
			{
				// When element exist
				return mid;
			}
			if (arr[low] < arr[mid])
			{
				// When elements are sort in location low to mid position
				if (k >= arr[low] && k < arr[mid])
				{
					// When the element is possible in the (low) to (mid-1) position
					return self.binarySearch(arr, k, low, mid - 1);
				}
				else
				{
					// When the element is possible in the (mid+1) to (high) position
					return self.binarySearch(arr, k, mid + 1, high);
				}
			}
			else if (arr[mid] < arr[high])
			{
				if (k > arr[mid] && k <= arr[high])
				{
					// When the element is possible in the (mid+1) to (high) position
					return self.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					// When the element is possible in the (low) to (mid-1) position
					return self.binarySearch(arr, k, low, mid - 1);
				}
			}
			else if (arr[low] == arr[mid])
			{
				// When repeated array element exists
				if (arr[mid] != arr[high])
				{
					return self.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					// Check if that element exists in left subtree
					var result: Int = self.binarySearch(arr, k, low, mid - 1);
					if (result == -1)
					{
						// When element not exist in left subtree then
						// Check that element exists in right subtree
						result = self.binarySearch(arr, k, mid + 1, high);
					}
					return result;
				}
			}
		}
		return -1;
	}
	// Handle request to find array elements
	func findElement(_ arr: [Int], _ size: Int, _ k: Int)
	{
		let location: Int = self.binarySearch(arr, k, 0, size - 1);
		if (location == -1)
		{
			print("\n Element ", k ," Are not exist ", terminator: "");
		}
		else
		{
			print("\n Element ", k ," exist in location ", location, terminator: "");
		}
	}
}
func main()
{
	let task: SearchElement = SearchElement();
	// Defining sorted and rotated array of integer element
	let arr1: [Int] = [7, 8, 9, 1, 2, 3, 4, 5, 6];
	let arr2: [Int] = [11, 22, 43, 45, 51, 62, 73, 1, 9, 10];
	// Get the size
	var size: Int = arr1.count;
	task.display(arr1, size);
	// Test Cases
	task.findElement(arr1, size, 9);
	task.findElement(arr1, size, 3);
	// Get the size
	size = arr2.count;
	task.display(arr2, size);
	// Test Cases
	task.findElement(arr2, size, 9);
	task.findElement(arr2, size, 11);
	task.findElement(arr2, size, 67);
}
main();

Output

 Array Elements
 [    7    8    9    1    2    3    4    5    6 ]
 Element  9  exist in location  2
 Element  3  exist in location  5

 Array Elements
 [    11    22    43    45    51    62    73    1    9    10 ]
 Element  9  exist in location  8
 Element  11  exist in location  0
 Element  67  Are not exist
/*
    Kotlin program 
    Find an element in a sorted and rotated array
*/
class SearchElement
{
	//   Display array elements
	fun display(arr: Array<Int>, size: Int): Unit
	{
		print("\n\n Array Elements \n [");
		var i: Int = 0;
		while (i<size)
		{
			print("   " + arr[i]);
			i += 1;
		}
		print(" ]");
	}
	//   Finding the element location of sorted rotated array
	fun binarySearch(arr: Array<Int>, k: Int, low: Int, high: Int): Int
	{
		if (low>high)
		{
			return -1;
		}
		else
		{
			var mid: Int = (low + high) / 2;
			if (arr[mid] == k)
			{
				// When element exist
				return mid;
			}
			if (arr[low]<arr[mid])
			{
				// When elements are sort in location low to mid position
				if (k>= arr[low] && k<arr[mid])
				{
					// When the element is possible in the (low) to (mid-1) position
					return this.binarySearch(arr, k, low, mid - 1);
				}
				else
				{
					// When the element is possible in the (mid+1) to (high) position
					return this.binarySearch(arr, k, mid + 1, high);
				}
			}
			else
			if (arr[mid]<arr[high])
			{
				if (k>arr[mid] && k <= arr[high])
				{
					// When the element is possible in the (mid+1) to (high) position
					return this.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					// When the element is possible in the (low) to (mid-1) position
					return this.binarySearch(arr, k, low, mid - 1);
				}
			}
			else
			if (arr[low] == arr[mid])
			{
				// When repeated array element exists
				if (arr[mid] != arr[high])
				{
					return this.binarySearch(arr, k, mid + 1, high);
				}
				else
				{
					// Check if that element exists in left subtree
					var result: Int = this.binarySearch(arr, k, low, mid - 1);
					if (result == -1)
					{
						// When element not exist in left subtree then
						// Check that element exists in right subtree
						result = this.binarySearch(arr, k, mid + 1, high);
					}
					return result;
				}
			}
		}
		return -1;
	}
	// Handle request to find array elements
	fun findElement(arr: Array<Int>, size: Int, k: Int): Unit
	{
		var location: Int = this.binarySearch(arr, k, 0, size - 1);
		if (location == -1)
		{
			print("\n Element " + k + " Are not exist ");
		}
		else
		{
			print("\n Element " + k + " exist in location " + location);
		}
	}
}
fun main(args: Array<String>): Unit
{
	var task: SearchElement = SearchElement();
	// Defining sorted and rotated array of integer element
	var arr1: Array<Int> = arrayOf(7, 8, 9, 1, 2, 3, 4, 5, 6);
	var arr2: Array<Int> = arrayOf(11, 22, 43, 45, 51, 62, 73, 1, 9, 10);
	// Get the size
	var size: Int = arr1.count();
	task.display(arr1, size);
	// Test Cases
	task.findElement(arr1, size, 9);
	task.findElement(arr1, size, 3);
	// Get the size
	size = arr2.count();
	task.display(arr2, size);
	// Test Cases
	task.findElement(arr2, size, 9);
	task.findElement(arr2, size, 11);
	task.findElement(arr2, size, 67);
}

Output

 Array Elements
 [   7   8   9   1   2   3   4   5   6 ]
 Element 9 exist in location 2
 Element 3 exist in location 5

 Array Elements
 [   11   22   43   45   51   62   73   1   9   10 ]
 Element 9 exist in location 8
 Element 11 exist in location 0
 Element 67 Are not exist

Resultant Output Explanation

The given C program applies the binary search algorithm on two example arrays, arr1 and arr2, and finds specific target elements. It then displays the results.

In the output, for arr1, the algorithm successfully finds elements 9 and 3 at indices 2 and 5, respectively. For arr2, the algorithm finds elements 9 and 11 at indices 8 and 0, respectively, while element 67 is not present in the array.

Time Complexity

The time complexity of the binary search algorithm in the given code is O(log n) since it reduces the search space by half in each recursive call. The algorithm performs binary search recursively on the left or right half of the array based on the comparisons, making it efficient for sorted and rotated arrays.

In conclusion, the given algorithm effectively solves the problem of finding an element in a sorted and rotated array using a modified binary search approach. It has a time complexity of O(log n) and can efficiently handle arrays with repeated elements.

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