Print all triplets in sorted array that form AP
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++
#include <bits/stdc++.h>
using namespace std;
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++)
{
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]);
}
}
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
import java.io.*;
import java.util.*;
class GFG
{
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++)
{
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]);
}
}
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);
}
}
|
Python3
def printAllAPTriplets(arr, n) :
s = [];
for i in range ( 0 , n - 1 ) :
for j in range (i + 1 , n) :
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]);
arr = [ 2 , 6 , 9 , 12 , 17 ,
22 , 31 , 32 , 35 , 42 ];
n = len (arr);
printAllAPTriplets(arr, n);
|
C#
using System;
using System.Collections.Generic;
class GFG
{
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++)
{
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]);
}
}
static void Main()
{
int []arr = new int []{ 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 };
int n = arr.Length;
printAllAPTriplets(arr, n);
}
}
|
PHP
<?php
function printAllAPTriplets( $arr , $n )
{
$s = array ();
for ( $i = 0; $i < $n - 1; $i ++)
{
for ( $j = $i + 1;
$j < $n ; $j ++)
{
$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 ]);
}
}
$arr = array (2, 6, 9, 12, 17,
22, 31, 32, 35, 42);
$n = count ( $arr );
printAllAPTriplets( $arr , $n );
?>
|
Javascript
<script>
function printAllAPTriplets( arr, n){
const s = new Set()
for (let i = 0; i < n - 1; i++)
{
for (let j = i + 1; j < n; j++)
{
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]);
}
}
let arr = [ 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 ];
let n = arr.length;
printAllAPTriplets(arr, n);
</script>
|
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.
C++
#include <bits/stdc++.h>
using namespace std;
void printAllAPTriplets( int arr[], int n)
{
for ( int i = 1; i < n - 1; i++)
{
for ( int j = i - 1, k = i + 1; j >= 0 && k < n;)
{
if (arr[j] + arr[k] == 2 * arr[i])
{
cout << arr[j] << " " << arr[i]
<< " " << arr[k] << endl;
k++;
j--;
}
else if (arr[j] + arr[k] < 2 * arr[i])
k++;
else
j--;
}
}
}
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
import java.io.*;
class GFG
{
static void findAllTriplets( int arr[], int n)
{
for ( int i = 1 ; i < n - 1 ; i++)
{
for ( int j = i - 1 , k = i + 1 ; j >= 0 && k < n;)
{
if (arr[j] + arr[k] == 2 * arr[i])
{
System.out.println(arr[j] + " " +
arr[i]+ " " + arr[k]);
k++;
j--;
}
else if (arr[j] + arr[k] < 2 * arr[i])
k++;
else
j--;
}
}
}
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);
}
}
|
Python 3
def printAllAPTriplets(arr, n):
for i in range ( 1 , n - 1 ):
j = i - 1
k = i + 1
while (j > = 0 and k < n ):
if (arr[j] + arr[k] = = 2 * arr[i]):
print (arr[j], " ", arr[i], " ", arr[k])
k + = 1
j - = 1
elif (arr[j] + arr[k] < 2 * arr[i]):
k + = 1
else :
j - = 1
arr = [ 2 , 6 , 9 , 12 , 17 ,
22 , 31 , 32 , 35 , 42 ]
n = len (arr)
printAllAPTriplets(arr, n)
|
C#
using System;
class GFG
{
static void findAllTriplets( int []arr, int n)
{
for ( int i = 1; i < n - 1; i++)
{
for ( int j = i - 1, k = i + 1; j >= 0 && k < n;)
{
if (arr[j] + arr[k] == 2 * arr[i])
{
Console.WriteLine(arr[j] + " " +
arr[i]+ " " + arr[k]);
k++;
j--;
}
else if (arr[j] + arr[k] < 2 * arr[i])
k++;
else
j--;
}
}
}
public static void Main ()
{
int []arr = { 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 };
int n = arr.Length;
findAllTriplets(arr, n);
}
}
|
PHP
<?php
function findAllTriplets( $arr , $n )
{
for ( $i = 1; $i < $n - 1; $i ++)
{
for ( $j = $i - 1, $k = $i + 1;
$j >= 0 && $k < $n 😉
{
if ( $arr [ $j ] + $arr [ $k ] == 2 *
$arr [ $i ])
{
echo $arr [ $j ] . " " .
$arr [ $i ]. " " .
$arr [ $k ] . "\n" ;
$k ++;
$j --;
}
else if ( $arr [ $j ] + $arr [ $k ] < 2 *
$arr [ $i ])
$k ++;
else
$j --;
}
}
}
$arr = array (2, 6, 9, 12, 17,
22, 31, 32, 35, 42);
$n = count ( $arr );
findAllTriplets( $arr , $n );
?>
|
Javascript
<script>
function printAllAPTriplets(arr, n)
{
for (let i = 1; i < n - 1; i++)
{
for (let j = i - 1, k = i + 1; j >= 0 && k < n;)
{
if (arr[j] + arr[k] == 2 * arr[i])
{
document.write(arr[j] + " " + arr[i]
+ " " + arr[k] + "<br>" );
k++;
j--;
}
else if (arr[j] + arr[k] < 2 * arr[i])
k++;
else
j--;
}
}
}
let arr = [ 2, 6, 9, 12, 17,
22, 31, 32, 35, 42 ];
let n = arr.length;
printAllAPTriplets(arr, n);
</script>
|
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)
Last Updated :
04 Aug, 2022
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