Given a sorted and rotated array, find if there is a pair with a given sum
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 have discussed a O(n) solution for a sorted array (See steps 2, 3 and 4 of Method 1). 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 (as discussed here in method 1) 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++
// 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; } |
Java
// Java program to find a pair with a given // sum in a sorted and rotated array class PairInSortedRotated { // This function returns true if arr[0..n-1] // has a pair with sum equals to x. static boolean 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 */ public static void main (String[] args) { int arr[] = { 11 , 15 , 6 , 8 , 9 , 10 }; int sum = 16 ; int n = arr.length; if (pairInSortedRotated(arr, n, sum)) System.out.print( "Array has two elements" + " with sum 16" ); else System.out.print( "Array doesn't have two" + " elements with sum 16 " ); } } /*This code is contributed by Prakriti Gupta*/ |
Python3
# 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 " ) # This article contributed by saloni1297 |
C#
// C# program to find a pair with a given // sum in a sorted and rotated array using System; class PairInSortedRotated { // This function returns true if arr[0..n-1] // has a pair with sum equals to x. static 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 ; // l is now index of smallest element int l = (i + 1) % n; // r is now index of largest element int 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; // Move to the lower sum side else r = (n + r - 1) % n; } return false ; } // Driver Code public static void Main () { int []arr = {11, 15, 6, 8, 9, 10}; int sum = 16; int n = arr.Length; if (pairInSortedRotated(arr, n, sum)) Console.WriteLine( "Array has two elements" + " with sum 16" ); else Console.WriteLine( "Array doesn't have two" + " elements with sum 16 " ); } } // This code is contributed by vt_m. |
PHP
<?php // PHP 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. function pairInSortedRotated( $arr , $n , $x ) { // Find the pivot element $i ; for ( $i = 0; $i < $n - 1; $i ++) 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 ; // Move to the lower sum side else $r = ( $n + $r - 1) % $n ; } return false; } // Driver Code $arr = array (11, 15, 6, 8, 9, 10); $sum = 16; $n = sizeof( $arr ); if (pairInSortedRotated( $arr , $n , $sum )) echo "Array has two elements " . "with sum 16" ; else echo "Array doesn't have two " . "elements with sum 16 " ; // This code is contributed by aj_36 ?> |
Javascript
<script> // Javascript 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. function pairInSortedRotated(arr, n, x) { // Find the pivot element let i; for (i = 0; i < n - 1; i++) if (arr[i] > arr[i + 1]) break ; // l is now index of // smallest element let l = (i + 1) % n; // r is now index of largest // element let 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; // Move to the lower sum side else r = (n + r - 1) % n; } return false ; } // Driver code let arr = [ 11, 15, 6, 8, 9, 10 ]; let sum = 16; let n = arr.length; if (pairInSortedRotated(arr, n, sum)) document.write( "Array has two elements" + " with sum 16" ); else document.write( "Array doesn't have two" + " elements with sum 16 " ); // This code is contributed by avanitrachhadiya2155 </script> |
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:
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; } |
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 sum = 16 ; int n = arr.length; System.out.println( pairsInSortedRotated(arr, n, sum)); } } // 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 ] s = 16 print (pairsInSortedRotated(arr, 6 , s)) # 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 sum = 16; int n = arr.Length; Console.WriteLine( pairsInSortedRotated(arr, n, sum)); } } // This code is contributed by akt_mit |
PHP
<?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); $sum = 16; $n = sizeof( $arr ) / sizeof( $arr [0]); echo pairsInSortedRotated( $arr , $n , $sum ); // This code is contributed by ajit ?> |
Javascript
<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 sum = 16; let n = arr.length; document.write(pairsInSortedRotated(arr, n, sum)); // This code is contributed by rag2127 </script> |
Output:
2
Time Complexity: O(n)
Auxiliary Space: O(1)
This method is suggested by Nikhil Jindal.
Exercise:
1) Extend the above solution to work for arrays with duplicates allowed.
This article is contributed by Himanshu Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above