Print all triplets in sorted array that form AP
Last Updated :
04 Aug, 2022
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.
Implementation:
C++
// C++ program to print all triplets in given
// array that form Arithmetic Progression
// C++ program to print all triplets in given
// array that form Arithmetic Progression
#include <bits/stdc++.h>
using namespace std;
// Function to print all triplets in
// given sorted array that forms AP
void printAllAPTriplets(int arr[], int n)
{
unordered_set<int> s;
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];
if (s.find(arr[i] - diff) != s.end())
cout << arr[i] - diff << " " << arr[i]
<< " " << arr[j] << endl;
}
s.insert(arr[i]);
}
}
// Driver code
int main()
{
int arr[] = { 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 };
int n = sizeof(arr) / sizeof(arr[0]);
printAllAPTriplets(arr, n);
return 0;
}
// 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)
# Python program to print all
# triplets in given array
# that form Arithmetic
# Progression
# Function to print
# all triplets in
# given sorted array
# that forms AP
def printAllAPTriplets(arr, n) :
s = [];
for i in range(0, n - 1) :
for j in range(i + 1, n) :
# Use hash to find if
# there is a previous
# element with difference
# equal to arr[j] - arr[i]
diff = arr[j] - arr[i];
if ((arr[i] - diff) in arr) :
print ("{} {} {}" .
format((arr[i] - diff),
arr[i], arr[j]),
end = "\n");
s.append(arr[i]);
# Driver code
arr = [2, 6, 9, 12, 17,
22, 31, 32, 35, 42];
n = len(arr);
printAllAPTriplets(arr, n);
# This code is contributed by
# Manish Shaw(manishshaw1)
// C# program to print all
// triplets in given array
// that form Arithmetic
// Progression
using System;
using System.Collections.Generic;
class GFG
{
// Function to print
// all triplets in
// given sorted array
// that forms AP
static void printAllAPTriplets(int []arr,
int n)
{
List<int> s = new List<int>();
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];
bool exists = s.Exists(element =>
element == (arr[i] -
diff));
if (exists)
Console.WriteLine(arr[i] - diff +
" " + arr[i] +
" " + arr[j]);
}
s.Add(arr[i]);
}
}
// Driver code
static void Main()
{
int []arr = new int[]{ 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)
<?php
// PHP program to pr$all
// triplets in given array
// that form Arithmetic
// Progression
// Function to print
// all triplets in
// given sorted array
// that forms AP
function printAllAPTriplets($arr, $n)
{
$s = array();
for ($i = 0; $i < $n - 1; $i++)
{
for ($j = $i + 1;
$j < $n; $j++)
{
// Use hash to find if
// there is a previous
// element with difference
// equal to arr[j] - arr[i]
$diff = $arr[$j] - $arr[$i];
if (in_array($arr[$i] -
$diff, $arr))
echo(($arr[$i] - $diff) .
" " . $arr[$i] .
" " . $arr[$j] . "\n");
}
array_push($s, $arr[$i]);
}
}
// Driver code
$arr = array(2, 6, 9, 12, 17,
22, 31, 32, 35, 42);
$n = count($arr);
printAllAPTriplets($arr, $n);
// This code is contributed by
// Manish Shaw(manishshaw1)
?>
<script>
// JavaScript program to print all triplets in given
// array that form Arithmetic Progression
// Function to print all triplets in
// given sorted array that forms AP
function printAllAPTriplets( arr, n){
const s = new Set()
for (let i = 0; i < n - 1; i++)
{
for (let j = i + 1; j < n; j++)
{
// Use hash to find if there is
// a previous element with difference
// equal to arr[j] - arr[i]
let diff = arr[j] - arr[i];
if (s.has(arr[i] - diff))
document.write(arr[i] - diff +" " + arr[i]
+ " " + arr[j] + "<br>");
}
s.add(arr[i]);
}
}
// Driver code
let arr = [ 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 ];
let n = arr.length;
printAllAPTriplets(arr, n);
</script>
Output6 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.
C++
// C++ program to print all triplets in given
// array that form Arithmetic Progression
#include <bits/stdc++.h>
using namespace std;
// Function to print all triplets in
// given sorted array that forms AP
void printAllAPTriplets(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])
{
cout << arr[j] << " " << arr[i]
<< " " << arr[k] << endl;
// 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
int main()
{
int arr[] = { 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 };
int n = sizeof(arr) / sizeof(arr[0]);
printAllAPTriplets(arr, n);
return 0;
}
// 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.
# python 3 program to print all triplets in given
# array that form Arithmetic Progression
# Function to print all triplets in
# given sorted array that forms AP
def printAllAPTriplets(arr, n):
for i in range(1, n - 1):
# Search other two elements of
# AP with arr[i] as middle.
j = i - 1
k = i + 1
while(j >= 0 and k < n ):
# if a triplet is found
if (arr[j] + arr[k] == 2 * arr[i]):
print(arr[j], "", arr[i], "", arr[k])
# Since elements are distinct,
# arr[k] and arr[j] cannot form
# any more triplets with arr[i]
k += 1
j -= 1
# If middle element is more move to
# higher side, else move lower side.
elif (arr[j] + arr[k] < 2 * arr[i]):
k += 1
else:
j -= 1
# Driver code
arr = [ 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 ]
n = len(arr)
printAllAPTriplets(arr, n)
# This article is contributed
# by Smitha Dinesh Semwal
// C# implementation to print
// all the triplets in given array
// that form Arithmetic Progression
using System;
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])
{
Console.WriteLine(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 ()
{
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.
<?php
// PHP implementation to print
// all the triplets in given array
// that form Arithmetic Progression
// Function to print all triplets in
// given sorted array that forms AP
function findAllTriplets($arr, $n)
{
for ($i = 1; $i < $n - 1; $i++)
{
// Search other two elements
// of AP with arr[i] as middle.
for ($j = $i - 1, $k = $i + 1;
$j >= 0 && $k < $n;)
{
// if a triplet is found
if ($arr[$j] + $arr[$k] == 2 *
$arr[$i])
{
echo $arr[$j] ." " .
$arr[$i]. " " .
$arr[$k] . "\n";
// 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
$arr = array(2, 6, 9, 12, 17,
22, 31, 32, 35, 42);
$n = count($arr);
findAllTriplets($arr, $n);
// This code is contributed by Sam007
?>
<script>
// JavaScript program to print all triplets in given
// array that form Arithmetic Progression
// Function to print all triplets in
// given sorted array that forms AP
function printAllAPTriplets(arr, n)
{
for (let i = 1; i < n - 1; i++)
{
// Search other two elements of
// AP with arr[i] as middle.
for (let j = i - 1, k = i + 1; j >= 0 && k < n;)
{
// if a triplet is found
if (arr[j] + arr[k] == 2 * arr[i])
{
document.write(arr[j] + " " + arr[i]
+ " " + arr[k] + "<br>");
// 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
let arr = [ 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 ];
let n = arr.length;
printAllAPTriplets(arr, n);
// This code is contributed by Surbhi Tyagi.
</script>
Output6 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)
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