Skip to content
Related Articles

Related Articles

Java Program to Print all triplets in sorted array that form AP

View Discussion
Improve Article
Save Article
  • Last Updated : 19 Jan, 2022
View Discussion
Improve Article
Save Article

Given a sorted array of distinct positive integers, print all triplets that form AP (or Arithmetic Progression)
Examples : 
 

Input : arr[] = { 2, 6, 9, 12, 17, 22, 31, 32, 35, 42 };
Output :
6 9 12
2 12 22
12 17 22
2 17 32
12 22 32
9 22 35
2 22 42
22 32 42

Input : arr[] = { 3, 5, 6, 7, 8, 10, 12};
Output :
3 5 7
5 6 7
6 7 8
6 8 10
8 10 12

 

A simple solution is to run three nested loops to generate all triplets and for every triplet, check if it forms AP or not. Time complexity of this solution is O(n3)
A better solution is to use hashing. We traverse array from left to right. We consider every element as middle and all elements after it as next element. To search the previous element, we use a hash table.
 

Java




// Java program to print all 
// triplets in given array 
// that form Arithmetic 
// Progression
import java.io.*;
import java.util.*;
  
class GFG
{
    // Function to print
    // all triplets in
    // given sorted array 
    // that forms AP
    static void printAllAPTriplets(int []arr,
                                   int n)
    {
        ArrayList<Integer> s = 
                 new ArrayList<Integer>();
        for (int i = 0;
                 i < n - 1; i++)
        {
            for (int j = i + 1; j < n; j++)
            {
                // Use hash to find if
                // there is a previous 
                // element with difference
                // equal to arr[j] - arr[i]
                int diff = arr[j] - arr[i];
                boolean exists = 
                        s.contains(arr[i] - diff);
                  
                if (exists)
                    System.out.println(arr[i] - diff + 
                                        " " + arr[i] + 
                                        " " + arr[j]);
            }
              
        s.add(arr[i]);
        }
    }
      
    // Driver code
    public static void main(String args[])
    {
        int []arr = {2, 6, 9, 12, 17
                     22, 31, 32, 35, 42};
        int n = arr.length;
        printAllAPTriplets(arr, n);
    }
}
  
// This code is contributed by 
// Manish Shaw(manishshaw1)


Output :  

6 9 12
2 12 22
12 17 22
2 17 32
12 22 32
9 22 35
2 22 42
22 32 42

Time Complexity : O(n2
Auxiliary Space : O(n)
An efficient solution is based on the fact that the array is sorted. We use the same concept as discussed in GP triplet question. The idea is to start from the second element and fix every element as a middle element and search for the other two elements in a triplet (one smaller and one greater). 
Below is the implementation of the above idea. 
 

Java




// Java implementation to print
// all the triplets in given array
// that form Arithmetic Progression
   
import java.io.*;
   
class GFG 
{
       
    // Function to print all triplets in
    // given sorted array that forms AP
    static void findAllTriplets(int arr[], int n)
    {
           
        for (int i = 1; i < n - 1; i++) 
        {
       
            // Search other two elements 
            // of AP with arr[i] as middle.
            for (int j = i - 1, k = i + 1; j >= 0 && k < n;)
            {
                   
                // if a triplet is found
                if (arr[j] + arr[k] == 2 * arr[i]) 
                {
                    System.out.println(arr[j] +" "
                                       arr[i]+ " " + arr[k]);
       
                    // Since elements are distinct,
                    // arr[k] and arr[j] cannot form
                    // any more triplets with arr[i]
                    k++;
                    j--;
                }
       
                // If middle element is more move to 
                // higher side, else move lower side.
                else if (arr[j] + arr[k] < 2 * arr[i]) 
                    k++;         
                else
                    j--;         
            }
        }
    }
   
    // Driver code
    public static void main (String[] args) 
    {
           
        int arr[] = { 2, 6, 9, 12, 17
                      22, 31, 32, 35, 42 };
        int n = arr.length;
           
        findAllTriplets(arr, n);
    }
}
   
// This code is contributed by vt_m.


Output :  

6 9 12
2 12 22
12 17 22
2 17 32
12 22 32
9 22 35
2 22 42
22 32 42

Time Complexity : O(n2
Auxiliary Space : O(1)
Please refer complete article on Print all triplets in sorted array that form AP for more details!


My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!