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++
#include <bits/stdc++.h>
using namespace std;
int pairsInSortedRotated( int arr[], int n, int x)
{
int i;
for (i = 0; i < n - 1; i++)
if (arr[i] > arr[i + 1])
break ;
int l = (i + 1) % n;
int r = i;
int cnt = 0;
while (l != r) {
if (arr[l] + arr[r] == x) {
cnt++;
if (l == (r - 1 + n) % n) {
return cnt;
}
l = (l + 1) % n;
r = (r - 1 + n) % n;
}
else if (arr[l] + arr[r] < x)
l = (l + 1) % n;
else
r = (n + r - 1) % n;
}
return cnt;
}
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
import java.io.*;
class GFG {
static int pairsInSortedRotated( int arr[], int n, int x)
{
int i;
for (i = 0 ; i < n - 1 ; i++)
if (arr[i] > arr[i + 1 ])
break ;
int l = (i + 1 ) % n;
int r = i;
int cnt = 0 ;
while (l != r) {
if (arr[l] + arr[r] == x) {
cnt++;
if (l == (r - 1 + n) % n) {
return cnt;
}
l = (l + 1 ) % n;
r = (r - 1 + n) % n;
}
else if (arr[l] + arr[r] < x)
l = (l + 1 ) % n;
else
r = (n + r - 1 ) % n;
}
return cnt;
}
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));
}
}
|
Python3
def pairsInSortedRotated(arr, n, x):
for i in range (n):
if arr[i] > arr[i + 1 ]:
break
l = (i + 1 ) % n
r = i
cnt = 0
while (l ! = r):
if arr[l] + arr[r] = = x:
cnt + = 1
if l = = (r - 1 + n) % n:
return cnt
l = (l + 1 ) % n
r = (r - 1 + n) % n
elif arr[l] + arr[r] < x:
l = (l + 1 ) % n
else :
r = (n + r - 1 ) % n
return cnt
arr = [ 11 , 15 , 6 , 7 , 9 , 10 ]
X = 16
N = len (arr)
print (pairsInSortedRotated(arr, N, X))
|
C#
using System;
class GFG {
static int pairsInSortedRotated( int [] arr, int n, int x)
{
int i;
for (i = 0; i < n - 1; i++)
if (arr[i] > arr[i + 1])
break ;
int l = (i + 1) % n;
int r = i;
int cnt = 0;
while (l != r) {
if (arr[l] + arr[r] == x) {
cnt++;
if (l == (r - 1 + n) % n) {
return cnt;
}
l = (l + 1) % n;
r = (r - 1 + n) % n;
}
else if (arr[l] + arr[r] < x)
l = (l + 1) % n;
else
r = (n + r - 1) % n;
}
return cnt;
}
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));
}
}
|
PHP
<?php
function pairsInSortedRotated( $arr ,
$n , $x )
{
$i ;
for ( $i = 0; $i < $n - 1; $i ++)
if ( $arr [ $i ] > $arr [ $i + 1])
break ;
$l = ( $i + 1) % $n ;
$r = $i ;
$cnt = 0;
while ( $l != $r )
{
if ( $arr [ $l ] + $arr [ $r ] == $x )
{
$cnt ++;
if ( $l == ( $r - 1 + $n ) % $n )
{
return $cnt ;
}
$l = ( $l + 1) % $n ;
$r = ( $r - 1 + $n ) % $n ;
}
else if ( $arr [ $l ] + $arr [ $r ] < $x )
$l = ( $l + 1) % $n ;
else
$r = ( $n + $r - 1) % $n ;
}
return $cnt ;
}
$arr = array (11, 15, 6,
7, 9, 10);
$X = 16;
$N = sizeof( $arr ) / sizeof( $arr [0]);
echo pairsInSortedRotated( $arr , $N , $X );
?>
|
Javascript
<script>
function pairsInSortedRotated(arr, n, x)
{
let i;
for (i = 0; i < n - 1; i++)
if (arr[i] > arr[i + 1])
break ;
let l = (i + 1) % n;
let r = i;
let cnt = 0;
while (l != r)
{
if (arr[l] + arr[r] == x)
{
cnt++;
if (l == (r - 1 + n) % n)
{
return cnt;
}
l = (l + 1) % n;
r = (r - 1 + n) % n;
}
else if (arr[l] + arr[r] < x)
l = (l + 1) % n;
else
r = (n + r - 1) % n;
}
return cnt;
}
let arr = [11, 15, 6, 7, 9, 10];
let X = 16;
let N = arr.length;
document.write(pairsInSortedRotated(arr, N, X));
</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.
Last Updated :
19 Sep, 2023
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