# Given a sorted and rotated array, find if there is a pair with a given sum

Given an array that is sorted and then rotated around an unknown point. Find if array has a pair with given sum ‘x’. It may be assumed that all elements in array are distinct.

Examples:

```Input: arr[] = {11, 15, 6, 8, 9, 10}, x = 16
Output: true
There is a pair (6, 10) with sum 16

Input: arr[] = {11, 15, 26, 38, 9, 10}, x = 35
Output: true
There is a pair (26, 9) with sum 35

Input: arr[] = {11, 15, 26, 38, 9, 10}, x = 45
Output: false
There is no pair with sum 45.
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

We have discussed a O(n) solution for a sorted array (See steps 2, 3 and 4 of Method 1). We can extend this solution for rotated array as well. The idea is to first find the maximum element in array which is the pivot point also and the element just before maximum is the minimum element. Once we have indexes maximum and minimum elements, we use similar meet in middle algorithm (as discussed here in method 1) to find if there is a pair. The only thing new here is indexes are incremented and decremented in rotational manner using modular arithmetic.

Following is the implementation of above idea.

## C++

```// C++ program to find a pair with a given sum in a sorted and
// rotated array
#include<iostream>
using namespace std;

// This function returns true if arr[0..n-1] has a pair
// with sum equals to x.
bool pairInSortedRotated(int arr[], int n, int x)
{
// Find the pivot element
int i;
for (i=0; i<n-1; i++)
if (arr[i] > arr[i+1])
break;
int l = (i+1)%n;  // l is now index of minimum element
int r = i;        // r is now index of maximum element

// Keep moving either l or r till they meet
while (l != r)
{
// If we find a pair with sum x, we return true
if (arr[l] + arr[r] == x)
return true;

// If current pair sum is less, move to the higher sum
if (arr[l] + arr[r] < x)
l = (l + 1)%n;
else  // Move to the lower sum side
r = (n + r - 1)%n;
}
return false;
}

/* Driver program to test above function */
int main()
{
int arr[] = {11, 15, 6, 8, 9, 10};
int sum = 16;
int n = sizeof(arr)/sizeof(arr[0]);

if (pairInSortedRotated(arr, n, sum))
cout << "Array has two elements with sum 16";
else
cout << "Array doesn't have two elements with sum 16 ";

return 0;
}
```

## Java

```// Java program to find a pair with a given
// sum in a sorted and rotated array
class PairInSortedRotated
{
// This function returns true if arr[0..n-1]
// has a pair with sum equals to x.
static boolean pairInSortedRotated(int arr[],
int n, int x)
{
// Find the pivot element
int i;
for (i = 0; i < n - 1; i++)
if (arr[i] > arr[i+1])
break;

int l = (i + 1) % n; // l is now index of
// minimum element

int r = i;       // r is now index of maximum
//element

// Keep moving either l or r till they meet
while (l != r)
{
// If we find a pair with sum x, we
// return true
if (arr[l] + arr[r] == x)
return true;

// If current pair sum is less, move
// to the higher sum
if (arr[l] + arr[r] < x)
l = (l + 1) % n;

else  // Move to the lower sum side
r = (n + r - 1) % n;
}
return false;
}

/* Driver program to test above function */
public static void main (String[] args)
{
int arr[] = {11, 15, 6, 8, 9, 10};
int sum = 16;
int n = arr.length;

if (pairInSortedRotated(arr, n, sum))
System.out.print("Array has two elements" +
" with sum 16");
else
System.out.print("Array doesn't have two" +
" elements with sum 16 ");
}
}
/*This code is contributed by Prakriti Gupta*/
```

Output:

`Array has two elements with sum 16`

Time complexity of the above solution is O(n). The step to find the pivot can be optimized to O(Logn) using the Binary Search approach discussed here.

Exercise:
1) Extend the above solution to work for arrays with duplicates allowed.

2) Extend the above solution to find all pairs.

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