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

**Examples :**

Input: arr[] = {11, 15, 6, 8, 9, 10}, x = 26 Output: true There is a pair (11, 15) with sum 26 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 = 52 Output: false There is no pair with sum 52.

We can extend this solution for rotated array as well. The idea is to first find the largest element in array which is the pivot point also and the element just after largest is the smallest element. Once we have indexes largest and smallest elements, we use similar meet in middle algorithm 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.

```
# Python3 program to find a
# pair with a given sum in
# a sorted and rotated array
# This function returns True
# if arr[0..n-1] has a pair
# with sum equals to x.
def pairInSortedRotated( arr, n, x ):
# Find the pivot element
for i in range(0, n - 1):
if (arr[i] > arr[i + 1]):
break;
# l is now index of smallest element
l = (i + 1) % n
# r is now index of largest element
r = i
# 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
arr = [11, 15, 6, 8, 9, 10]
sum = 16
n = len(arr)
if (pairInSortedRotated(arr, n, sum)):
print ("Array has two elements with sum 16")
else:
print ("Array doesn't have two elements with sum 16 ")
```

**Output : **

Array has two elements with sum 16

The 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.

**How to count all pairs having sum x?**

The stepwise algorithm is:

- Find the pivot element of the sorted and the rotated array. The pivot element is the largest element in the array. The smallest element will be adjacent to it.
- Use two pointers (say left and right) with the left pointer pointing to the smallest element and the right pointer pointing to largest element.
- Find the sum of the elements pointed by both the pointers.
- If the sum is equal to x, then increment the count. If the sum is less than x, then to increase sum move the left pointer to next position by incrementing it in a rotational manner. If the sum is greater than x, then to decrease sum move the right pointer to next position by decrementing it in rotational manner.
- Repeat step 3 and 4 until the left pointer is not equal to the right pointer or until the left pointer is not equal to right pointer – 1.
- Print final count.

Below is implementation of above algorithm:

```
# Python program to find
# number of pairs with
# a given sum in a sorted
# and rotated array.
# This function returns
# count of number of pairs
# with sum equals to x.
def pairsInSortedRotated(arr, n, x):
# Find the pivot element.
# Pivot element is largest
# element of array.
for i in range(n):
if arr[i] > arr[i + 1]:
break
# l is index of
# smallest element.
l = (i + 1) % n
# r is index of
# largest element.
r = i
# Variable to store
# count of number
# of pairs.
cnt = 0
# Find sum of pair
# formed by arr[l]
# and arr[r] and
# update l, r and
# cnt accordingly.
while (l != r):
# If we find a pair
# with sum x, then
# increment cnt, move
# l and r to next element.
if arr[l] + arr[r] == x:
cnt += 1
# This condition is
# required to be checked,
# otherwise l and r will
# cross each other and
# loop will never terminate.
if l == (r - 1 + n) % n:
return cnt
l = (l + 1) % n
r = (r - 1 + n) % n
# If current pair sum
# is less, move to
# the higher sum side.
elif arr[l] + arr[r] < x:
l = (l + 1) % n
# If current pair sum
# is greater, move to
# the lower sum side.
else:
r = (n + r - 1) % n
return cnt
# Driver Code
arr = [11, 15, 6, 7, 9, 10]
s = 16
print(pairsInSortedRotated(arr, 6, s))
```

**Output:**

2

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

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