Given an integer n, the task is to find whether n can be written as sum of three consecutive integer. If yes, find the three consecutive integer, else print “-1”.
Examples:
Input: n = 6
Output: 1 2 3
Explanation: 6 = 1 + 2 + 3.Input: n = 7
Output: -1
Method 1: (Brute Force):
The idea is to run a loop from i = 0 to n – 2, check if (i + i+1 + i+2) is equal to n. Also, check if n is positive or negative and accordingly increment or decrement i by 1.
Below is the implementation of this approach:
// CPP Program to check if a number can // be written as sum of three consecutive // integers. #include <bits/stdc++.h> using namespace std;
// function to check if a number can be written as sum of // three consecutive integer. void checksum( int n)
{ // if n is 0
if (n == 0) {
cout << "-1 0 1" << endl;
return ;
}
int inc;
// if n is positive, increment loop by 1.
if (n > 0)
inc = 1;
// if n is negative, decrement loop by 1.
else
inc = -1;
// Running loop from 0 to n - 2
for ( int i = 0; i <= n - 2; i += inc) {
// check if sum of three consecutive
// integer is equal to n.
if (i + i + 1 + i + 2 == n) {
cout << i << " " << i + 1
<< " " << i + 2;
return ;
}
}
cout << "-1" ;
} // Driver Program int main()
{ int n = 6;
checksum(n);
return 0;
} |
// JAVA Code to check if a number // can be written as sum of // three consecutive integers. import java.util.*;
class GFG
{ // function to check if a number
// can be written as sum of
// three consecutive integer.
static void checksum( int n)
{
// if n is 0
if (n == 0 ) {
System.out.println( "-1 0 1" );
return ;
}
int inc;
// if n is positive,
// increment loop by 1.
if (n > 0 )
inc = 1 ;
// if n is negative,
// decrement loop by 1.
else
inc = - 1 ;
// Running loop from 0 to n - 2
for ( int i = 0 ; i <= n - 2 ; i += inc) {
// check if sum of three consecutive
// integer is equal to n.
if (i + i + 1 + i + 2 == n) {
System.out.println(i + " " +
(i + 1 ) +
" " + (i + 2 ));
return ;
}
}
System.out.println( "-1" );
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 6 ;
checksum(n);
}
} // This code is contributed by Arnav Kr. Mandal. |
# Python3 code to check if a number # can be written as sum of three # consecutive integers. # function to check if a number # can be written as sum of three # consecutive integer. def checksum(n):
# if n is 0
if n = = 0 :
print ( "-1 0 1" )
return 0
inc = 0 # if n is positive,
# increment loop by 1.
if n > 0 :
inc = 1
# if n is negative,
# decrement loop by 1.
else :
inc = - 1
# Running loop from 0 to n - 2
for i in range ( 0 , n - 1 , inc):
# check if sum of three consecutive
# integer is equal to n.
if i + i + 1 + i + 2 = = n:
print (i , " " ,i + 1 , " " , i + 2 )
return 0
print ( "-1" )
# Driver Code n = 6
checksum(n) # This code is contributed by "Sharad_Bhardwaj". |
// C# Code to check if a number // can be written as sum of // three consecutive integers. using System;
class GFG
{ // function to check if a number
// can be written as sum of
// three consecutive integer.
static void checksum( int n)
{
// if n is 0
if (n == 0) {
Console.WriteLine( "-1 0 1" );
return ;
}
int inc;
// if n is positive,
// increment loop by 1.
if (n > 0)
inc = 1;
// if n is negative,
// decrement loop by 1.
else
inc = -1;
// Running loop from 0 to n - 2
for ( int i = 0; i <= n - 2; i += inc) {
// check if sum of three consecutive
// integer is equal to n.
if (i + i + 1 + i + 2 == n) {
Console.WriteLine(i + " "
+ (i + 1) + " " + (i + 2));
return ;
}
}
Console.WriteLine( "-1" );
}
/* Driver program to test above function */
public static void Main()
{
int n = 6;
checksum(n);
}
} // This code is contributed by vt_m. |
<?php // PHP Program to check if a // number can be written // as sum of three consecutive // integers. // function to check if a number // can be written as sum of // three consecutive integer. function checksum( $n )
{ // if n is 0
if ( $n == 0)
{
echo "-1 0 1" ;
return ;
}
$inc ;
// if n is positive,
// increment loop by 1.
if ( $n > 0)
$inc = 1;
// if n is negative,
// decrement loop by 1.
else
$inc = -1;
// Running loop from
// 0 to n - 2
for ( $i = 0; $i <= $n - 2; $i += $inc )
{
// check if sum of three consecutive
// integer is equal to n.
if ( $i + $i + 1 + $i + 2 == $n )
{
echo $i , " " , $i + 1
, " " , $i + 2;
return ;
}
}
echo "-1" ;
} // Driver Code
$n = 6;
checksum( $n );
// This code is contributed by anuj_67. ?> |
<script> // Javascript Code to check if a number // can be written as sum of // three consecutive integers. // function to check if a number
// can be written as sum of
// three consecutive integer.
function checksum(n)
{
// if n is 0
if (n == 0) {
document.write( "-1 0 1" );
return ;
}
var inc;
// if n is positive,
// increment loop by 1.
if (n > 0)
inc = 1;
// if n is negative,
// decrement loop by 1.
else
inc = -1;
// Running loop from 0 to n - 2
for (i = 0; i <= n - 2; i += inc)
{
// check if sum of three consecutive
// integer is equal to n.
if (i + i + 1 + i + 2 == n) {
document.write(i + " " + (i + 1) + " "
+ (i + 2));
return ;
}
}
document.write( "-1" );
}
/* Driver program to test above function */
var n = 6;
checksum(n);
// This code is contributed by gauravrajput1 </script> |
1 2 3
Time Complexity: O(N)
Auxiliary Space: O(1)
Method 2: (Efficient Approach)
The idea is to check if n is multiple of 3 or not.
Let n is sum of three consecutive integer of k – 1, k, k + 1. Therefore,
k – 1 + k + k + 1 = n
3*k = n
The three number will be n/3 – 1, n/3, n/3 + 1.
// CPP Program to check if a number can be // written as sum of three consecutive integer. #include <bits/stdc++.h> using namespace std;
// function to check if a number can be // written as sum of three consecutive // integers. void checksum( int n)
{ // if n is multiple of 3
if (n % 3 == 0)
cout << n / 3 - 1 << " " << n / 3 << " " << n / 3 + 1;
// else print "-1".
else
cout << "-1" ;
} // Driver Program int main()
{ int n = 6;
checksum(n);
return 0;
} |
// JAVA Code to check if a number // can be written as sum of three // consecutive integers. import java.util.*;
class GFG
{ // function to check if a number
// can be written as sum of three
// consecutive integers.
static void checksum( int n)
{
// if n is multiple of 3
if (n % 3 == 0 )
System.out.println( n / 3 - 1 + " "
+ n / 3 + " " + (n / 3 + 1 ));
// else print "-1".
else
System.out.println( "-1" );
}
/* Driver program to test above function */
public static void main(String[] args)
{
int n = 6 ;
checksum(n);
}
} // This code is contributed by Arnav Kr. Mandal. |
# Python3 code to check if a number # can be written as sum of three # consecutive integer. # function to check if a number # can be written as sum of three # consecutive integers. def checksum(n):
n = int (n)
# if n is multiple of 3
if n % 3 = = 0 :
print ( int (n / 3 - 1 ) , " " ,
int (n / 3 ), " " , int (n / 3 + 1 ))
# else print "-1".
else :
print ( "-1" )
# Driver Code n = 6
checksum(n) # This code is contributed by "Sharad_Bhardwaj". |
// C# Code to check if a number // can be written as sum of three // consecutive integers. using System;
class GFG
{ // function to check if a number
// can be written as sum of three
// consecutive integers.
static void checksum( int n)
{
// if n is multiple of 3
if (n % 3 == 0)
Console.WriteLine( n / 3 - 1 + " "
+ n / 3 + " " + (n / 3 + 1));
// else print "-1".
else
Console.WriteLine( "-1" );
}
/* Driver program to
test above function */
public static void Main()
{
int n = 6;
checksum(n);
}
} // This code is contributed by vt_m. |
<?php // PHP Code to check if a number // can be written as sum of three // consecutive integers. // function to check if // a number can be written // as sum of three consecutive // integers. function checksum( $n )
{ // if n is multiple of 3
if ( $n % 3 == 0)
echo $n / 3 - 1, " " ,
$n / 3, " " ,
$n / 3 + 1;
// else print "-1".
else
echo "-1" ;
} // Driver Program
$n = 6;
checksum( $n );
// This code is contributed by aj_36 ?> |
<script> // javascript Code to check if a number // can be written as sum of three // consecutive integers. // function to check if a number
// can be written as sum of three
// consecutive integers.
function checksum(n) {
// if n is multiple of 3
if (n % 3 == 0)
document.write(n / 3 - 1 + " " + n / 3 + " " + (n / 3 + 1));
// else print "-1".
else
document.write( "-1" );
}
/* Driver program to test above function */
var n = 6;
checksum(n);
// This code is contributed by todaysgaurav </script> |
1 2 3
Time Complexity: O(1)
Auxiliary Space: O(1)