Count of Pairs with given sum in Rotated Sorted Array
Last Updated :
23 Jul, 2025
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 <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 X = 16;
int N = sizeof(arr) / sizeof(arr[0]);
cout << pairsInSortedRotated(arr, N, X);
return 0;
}
// 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
# 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# 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
// PHP 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.
function pairsInSortedRotated($arr,
$n, $x)
{
// Find the pivot element.
// Pivot element is largest
// element of array.
$i;
for ($i = 0; $i < $n - 1; $i++)
if ($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)
{
$cnt++;
// This condition is required
// to be checked, otherwise l
// and r will cross each other
// and loop will never terminate.
if($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
?>
<script>
// Javascript 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.
function pairsInSortedRotated(arr, n, x)
{
// Find the pivot element.
// Pivot element is largest
// element of array.
let i;
for (i = 0; i < n - 1; i++)
if (arr[i] > arr[i + 1])
break;
// l is index of
// smallest element.
let l = (i + 1) % n;
// r is index of
// largest element.
let r = i;
// Variable to store
// count of number
// of pairs.
let 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
let arr = [11, 15, 6, 7, 9, 10];
let X = 16;
let N = arr.length;
document.write(pairsInSortedRotated(arr, N, X));
// This code is contributed by rag2127
</script>
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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