Check if a number can be written as sum of three consecutive integers

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: 

C++

// 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

// 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

# 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#

// 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
// 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.
?>

Javascript

<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>

Output

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. 

C++

// 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

// 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

# 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#

// 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
// 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
?>

Javascript

<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>

Output

1 2 3

Time Complexity: O(1)
Auxiliary Space: O(1)