Number expressed as sum of five consecutive integers
Given an integer n, the task is to find whether n can be expressed as sum of five consecutive integer. If yes, find the five consecutive integers, else print “-1”.
Examples:
Input : n = 15
Output : 1 2 3 4 5
15 = 1 + 2 + 3 + 4 + 5
Input : n = 18
Output : -1
Method 1: (Brute Force)
The idea is to run a loop from i = 0 to n – 4, check if (i + i+1 + i+2 + i+3 + i+4) 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:
C++
#include <bits/stdc++.h>
using namespace std;
void checksum( int n)
{
if (n == 0) {
cout << "-2 -1 0 1 2" << endl;
return ;
}
int inc;
if (n > 0)
inc = 1;
else
inc = -1;
for ( int i = 0; i <= n - 4; i += inc) {
if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) {
cout << i << " " << i + 1
<< " " << i + 2
<< " " << i + 3
<< " " << i + 4;
return ;
}
}
cout << "-1" ;
}
int main()
{
int n = 15;
checksum(n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void checksum( int n)
{
if (n == 0 ) {
System.out.println( "-2 -1 0 1 2" );
return ;
}
int inc;
if (n > 0 )
inc = 1 ;
else
inc = - 1 ;
for ( int i = 0 ; i <= n - 4 ; i += inc) {
if (i + i + 1 + i + 2 + i + 3 + i
+ 4 == n)
{
System.out.print( (i )
+ " " + (i + 1 )
+ " " + (i + 2 )
+ " " + (i + 3 )
+ " " + (i + 4 ));
return ;
}
}
System.out.println( "-1" );
}
public static void main (String[] args)
{
int n = 15 ;
checksum(n);
}
}
|
Python3
def checksum(n):
if n = = 0 :
print ( "-2 -1 0 1 2" )
return 0
inc = 0
if n > 0 :
inc = 1
else :
inc = - 1
for i in range ( 0 , n - 3 , inc):
if i + i + 1 + i + 2 + i + 3 + i + 4 = = n:
print (i, " " , i + 1 , " " , i + 2 , " " , i + 3 , " " , i + 4 )
return 0
print ( "-1" )
n = 15
checksum(n)
|
C#
using System;
class GFG {
static void checksum( int n)
{
if (n == 0) {
Console.Write( "-2 -1 0 1 2" );
return ;
}
int inc;
if (n > 0)
inc = 1;
else
inc = -1;
for ( int i = 0; i <= n - 4; i += inc) {
if (i + i + 1 + i + 2 + i + 3 + i
+ 4 == n)
{
Console.Write( (i )
+ " " + (i + 1)
+ " " + (i + 2)
+ " " + (i + 3)
+ " " + (i + 4));
return ;
}
}
Console.WriteLine( "-1" );
}
public static void Main ()
{
int n = 15;
checksum(n);
}
}
|
PHP
<?php
function checksum( $n )
{
if ( $n == 0)
{
echo "-2 -1 0 1 2" , "\n" ;
return ;
}
$inc ;
if ( $n > 0)
$inc = 1;
else
$inc = -1;
for ( $i = 0;
$i <= $n - 4; $i += $inc )
{
if ( $i + $i + 1 + $i + 2 +
$i + 3 + $i + 4 == $n )
{
echo $i , " " , $i + 1,
" " , $i + 2,
" " , $i + 3,
" " , $i + 4;
return ;
}
}
echo "-1" ;
}
$n = 15;
checksum( $n );
?>
|
Javascript
<script>
function checksum(n) {
if (n == 0) {
document.write( "-2 -1 0 1 2" );
return ;
}
var inc;
if (n > 0)
inc = 1;
else
inc = -1;
for (i = 0; i <= n - 4; i += inc) {
if (i + i + 1 + i + 2 + i + 3 + i + 4 == n) {
document.write((i) + " "
+ (i + 1) + " "
+ (i + 2) + " "
+ (i + 3) + " "
+ (i + 4));
return ;
}
}
document.write( "-1" );
}
var n = 15;
checksum(n);
</script>
|
Output :
1 2 3 4 5
Method 2: (Efficient Approach)
The idea is to check if n is multiple of 5 or not.
Let n is sum of five consecutive integer of k – 2, k-1, k, k + 1, k+2. Therefore,
k-2 + k-1 + k + k+1 + k+2= n
5*k = n
The five numbers will be n/5 – 2, n/5 – 1, n/5, n/5 + 1, n/5 + 2.
Below is the implementation of this approach:
C++
#include <bits/stdc++.h>
using namespace std;
void checksum( int n)
{
if (n % 5 == 0)
cout << n / 5 - 2 << " "
<< n / 5 - 1 << " " << n / 5
<< " " << n / 5 + 1 << " "
<< n / 5 + 2;
else
cout << "-1" ;
}
int main()
{
int n = 15;
checksum(n);
return 0;
}
|
Java
import java.io.*;
class GFG {
static void checksum( int n)
{
if (n % 5 == 0 )
System.out.println( (n / 5 - 2 )
+ " " + (n / 5 - 1 ) + " "
+ (n / 5 ) + " " + (n / 5
+ 1 ) + " " + (n / 5 + 2 ));
else
System.out.println( "-1" );
}
public static void main (String[] args)
{
int n = 15 ;
checksum(n);
}
}
|
Python3
def checksum(n):
n = int (n)
if n % 5 = = 0 :
print ( int (n / 5 - 2 ), " " ,
int (n / 5 - 1 ), " " , int (n / 5 ), " " , int (n / 5 + 1 ), " " , int (n / 5 + 2 ))
else :
print ( "-1" )
n = 15
checksum(n)
|
C#
using System;
class GFG {
static void checksum( int n)
{
if (n % 5 == 0)
Console.WriteLine( (n / 5 - 2)
+ " " + (n / 5 - 1) + " "
+ (n / 5) + " " + (n / 5
+ 1 ) + " " + (n / 5 + 2));
else
Console.WriteLine( "-1" );
}
public static void Main ()
{
int n = 15;
checksum(n);
}
}
|
PHP
<?php
function checksum( $n )
{
if ( $n % 5 == 0)
echo $n / 5 - 2 , " " ,
$n / 5 - 1 , " " ,
$n / 5 , " " ,
$n / 5 + 1 , " " ,
$n / 5 + 2;
else
echo "-1" ;
}
$n = 15;
checksum( $n );
?>
|
Javascript
<script>
function checksum(n) {
if (n % 5 == 0)
document.write((n / 5 - 2) + " "
+ (n / 5 - 1) + " " + (n / 5)
+ " " + (n / 5 + 1) + " " + (n / 5 + 2));
else
document.write( "-1" );
}
var n = 15;
checksum(n);
</script>
|
Output:
1 2 3 4 5
Time Complexity : O(1)
Auxilitary Space Complexity : O(1)
Last Updated :
06 Apr, 2023
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