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# Count of Pairs with given sum in Rotated Sorted Array

Given an array arr[] of distinct elements size N that is sorted and then around an unknown point, the task is to count the number of pairs in the array having a given sum X.

Examples:

Input: arr[] = {11, 15, 26, 38, 9, 10}, X = 35
Output: 1
Explanation: There is a pair (26, 9) with sum 35

Input: arr[] = {11, 15, 6, 7, 9, 10}, X = 16
Output: 2

Approach: The idea is similar to what is mentioned below.

First find the largest element in an array which is the pivot point also and the element just after the largest is the smallest element. Once we have the indices of the largest and the smallest elements, we use a similar meet-in-middle algorithm (as discussed here in method 1) to count the number of pairs that sum up to X. Indices are incremented and decremented in a rotational manner using modular arithmetic.

Follow the below illustration for a better understanding.

Illustration:

Let us take an example arr[]={11, 15, 6, 7, 9, 10}, X = 16, count=0;
Initially pivot = 1,

l = 2, r = 1:
=> arr[2] + arr[1] = 6 + 15 = 21, which is > 16
=> So decrement r = ( 6 + 1 – 1) % 6, r = 0

l = 2, r = 0:
=> arr[2] + arr[0] = 17 which is > 16,
=> So decrement r = (6 + 0 – 1) % 6, r = 5

l = 2, r = 5:
=> arr[2] + arr[5] = 16 which is equal to 16,
=> Hence count = 1 and
=> Decrement r = (6 + 5 – 1) % 6, r = 4 and increment l = (2 + 1) % 6, l = 3

l = 3, r = 4:
=> arr[3] + arr[4] = 16
=> Hence increment count. So count = 2
=> So decrement r = (6 + 4 – 1) % 6, r = 3 and increment l = 4

l = 4, r = 3:
=> l > r. So break the loop.

So we get count = 2

Follow the below steps to implement the idea:

• We will run a for loop from 0 to N-1, to find out the pivot point. Set the left pointer(l) to the smallest value and
the right pointer(r) to the highest value.
• To restrict the circular movement within the array, apply the modulo operation by the size of the array.
• While l ! = r, keep checking if arr[l] + arr[r] = sum.
• If arr[l] + arr[r] > sum, update r=(N+r-1) % N.
• If arr[l] + arr[r] < sum, update l=(l+1) % N.
• If arr[l] + arr[r] = sum, increment count. Also increment l and decrement r.

Below is the implementation of the above idea.

## C++

 `// C++ program to find number of pairs with``// a given sum in a sorted and rotated array.``#include ``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` `X = 16;``    ``int` `N = ``sizeof``(arr) / ``sizeof``(arr[0]);` `    ``cout << pairsInSortedRotated(arr, N, X);` `    ``return` `0;``}`

## Java

 `// Java program to find``// number of pairs with``// a given sum in a sorted``// and rotated array.``import` `java.io.*;` `class` `GFG {` `    ``// This function returns``    ``// count of number of pairs``    ``// with sum equals to x.``    ``static` `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 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 Code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``int` `arr[] = { ``11``, ``15``, ``6``, ``7``, ``9``, ``10` `};``        ``int` `X = ``16``;``        ``int` `N = arr.length;` `        ``System.out.println(pairsInSortedRotated(arr, N, X));``    ``}``}` `// This code is contributed by ajit`

## Python3

 `# 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``]``X ``=` `16``N ``=` `len``(arr)` `print``(pairsInSortedRotated(arr, N, X))` `# This code is contributed by ChitraNayal`

## C#

 `// C# program to find``// number of pairs with``// a given sum in a sorted``// and rotated array.``using` `System;` `class` `GFG {` `    ``// This function returns``    ``// count of number of pairs``    ``// with sum equals to x.``    ``static` `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 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 Code``    ``static` `public` `void` `Main()``    ``{``        ``int``[] arr = { 11, 15, 6, 7, 9, 10 };``        ``int` `X = 16;``        ``int` `N = arr.Length;` `        ``Console.WriteLine(``            ``pairsInSortedRotated(arr, N, X));``    ``}``}` `// This code is contributed by akt_mit`

## PHP

 ` ``\$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``++;``            ` `            ``// 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 Code``\$arr` `= ``array``(11, 15, 6,``              ``7, 9, 10);``\$X` `= 16;``\$N` `= sizeof(``\$arr``) / sizeof(``\$arr``[0]);` `echo` `pairsInSortedRotated(``\$arr``, ``\$N``, ``\$X``);` `// This code is contributed by ajit``?>`

## Javascript

 ``

Output

`2`

Time Complexity: O(N). As we are performing linear operations on an array.
Auxiliary Space: O(1). As constant extra space is used.

This method is suggested by Nikhil Jindal.

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