Hello Everyone,

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 = 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.

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 = 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.

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.

`// 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 smallest element`

` `

`int`

`r = i; `

`// r is now index of largest 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;`

`}`

**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.

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

The stepwise algo 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:

- C++

`// C++ program to find number of pairs with`

`// a given sum in a sorted and rotated array.`

`#include<bits/stdc++.h>`

`using`

`namespace`

`std;`

`// This function returns count of number of pairs`

`// with sum equals to x.`

`int`

`pairsInSortedRotated(`

`int`

`arr[], `

`int`

`n, `

`int`

`x)`

`{`

` `

`// Find the pivot element. Pivot element`

` `

`// is largest element of array.`

` `

`int`

`i;`

` `

`for`

`(i = 0; i < n-1; i++)`

` `

`if`

`(arr[i] > arr[i+1])`

` `

`break`

`;`

` `

` `

`// l is index of smallest element.`

` `

`int`

`l = (i + 1) % n;`

` `

` `

`// r is index of largest element.`

` `

`int`

`r = i;`

` `

` `

`// Variable to store count of number`

` `

`// of pairs.`

` `

`int`

`cnt = 0;`

` `

`// Find sum of pair formed by arr[l] and`

` `

`// 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++;`

` `

` `

`// 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.`

` `

`else`

`if`

`(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 program to test above function */`

`int`

`main()`

`{`

` `

`int`

`arr[] = {11, 15, 6, 7, 9, 10};`

` `

`int`

`sum = 16;`

` `

`int`

`n = `

`sizeof`

`(arr)/`

`sizeof`

`(arr[0]);`

` `

`cout << pairsInSortedRotated(arr, n, sum);`

` `

` `

`return`

`0;`

`}`

**Output:**

2

**Time Complexity:** O(n)

**Auxiliary Space:** O(1)