Java Program for Given a sorted and rotated array, find if there is a pair with a given sum

// 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*/

// 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