# Find an element in a sorted and rotated array

Here given code implementation process.

``````// 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``````

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