Break the number into three parts

Given a really large number, break it into 3 whole numbers such that they sum up to the original number and count the number of ways to do so.

Examples :

Input : 3
Output : 10
The possible combinations where the sum
of the numbers is equal to 3 are:
0+0+3 = 3
0+3+0 = 3
3+0+0 = 3
0+1+2 = 3
0+2+1 = 3
1+0+2 = 3
1+2+0 = 3
2+0+1 = 3
2+1+0 = 3
1+1+1 = 3

Input : 6
Output : 28

A total of 10 ways, so answer is 10. 

Naive Approach: Try all combinations from 0 to the given number and check if they add up to the given number or not, if they do, increase the count by 1 and continue the process. 

C++

// C++ program to count number of ways to break
// a number in three parts.
#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Function to count number of ways
// to make the given number n
ll count_of_ways(ll n)
{
    ll count = 0;
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= n; j++)
            for (int k = 0; k <= n; k++)
                if (i + j + k == n)
                    count++;
    return count;
}

// Driver Function
int main()
{
    ll n = 3;
    cout << count_of_ways(n) << endl;
    return 0;
}

Java

// Java program to count number of ways to break
// a number in three parts
import java.io.*;

class GFG {
    // Function to count number of ways
    // to make the given number n
    static long count_of_ways(long n)
    {
        long count = 0;
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n; j++)
                for (int k = 0; k <= n; k++)
                    if (i + j + k == n)
                        count++;
        return count;
    }

    // driver program
    public static void main(String[] args)
    {
        long n = 3;
        System.out.println(count_of_ways(n));
    }
}

// Contributed by Pramod Kumar

Python3

# Python3 program to count number of
# ways to break
# a number in three parts.

# Function to count number of ways
# to make the given number n
def count_of_ways(n):

    count = 0
    for i in range(0, n+1):
        for j in range(0, n+1):
            for k in range(0, n+1):
                if(i + j + k == n):
                    count = count + 1
    return count

# Driver Function
if __name__=='__main__':
    n = 3
    print(count_of_ways(n))


# This code is contributed by
# Sanjit_Prasad

C#

// C# program to count number of ways
// to break a number in three parts
using System;

class GFG {

    // Function to count number of ways
    // to make the given number n
    static long count_of_ways(long n)
    {
        long count = 0;
        for (int i = 0; i <= n; i++)
            for (int j = 0; j <= n; j++)
                for (int k = 0; k <= n; k++)
                    if (i + j + k == n)
                        count++;
        return count;
    }

    // driver program
    public static void Main()
    {
        long n = 3;
        Console.WriteLine(count_of_ways(n));
    }
}

// This code is Contributed by vt_m.

PHP

<?php
// PHP program to count number
// of ways to break a number
// in three parts.

// Function to count number of ways
// to make the given number n
function count_of_ways( $n)
{
    $count = 0;
    for ($i = 0; $i <= $n; $i++)
        for ($j = 0; $j <= $n; $j++)
            for ($k = 0; $k <= $n; $k++)
                if ($i + $j + $k == $n)
                    $count++;
    return $count;
}

// Driver Code
$n = 3;
echo count_of_ways($n);

// This code is Contributed by vt_m.
?>

Javascript

<script>

// JavaScript program to count
// number of ways to break
// a number in three parts.

// Function to count number of ways
// to make the given number n
function count_of_ways(n)
{
    let count = 0;
    for(let i = 0; i <= n; i++)
        for(let j = 0; j <= n; j++)
            for(let k = 0; k <= n; k++)
                if (i + j + k == n)
                    count++;

    return count;
}

// Driver code
let n = 3;

document.write(count_of_ways(n) + "<br>");

// This code is contributed by Surbhi Tyagi.

</script>

Output

10

Time Complexity : O(n³) 
Auxiliary Space: O(1) 

Another Approach:

The idea is to run two nested for loops to choose two numbers between 0 to n. After that, on subtracting both numbers from n, if we get any value greater than or equal to 0, then the third number exists that satisfies the required condition and it is that number. So increment the count.

Steps to implement-

• Initialize a variable count with a value of 0
• After that run two nested loops from 0 to n
• If after subtracting those numbers from n we get any value greater than or equal to 0. Then the third number exists that satisfies the question’s condition and it is that number.
• So in that case increment count by 1
• In the last print/return the count

Code-

C++

// C++ program to count number of ways to break
// a number in three parts.
#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Function to count number of ways
// to make the given number n
ll count_of_ways(ll n)
{
    ll count = 0;
    for (int i = 0; i <= n; i++)
        for (int j = 0; j <= n; j++)
        //Third number satisfying the input condition
        //will be n-(i+j) if n-(i+j)>=0
            if(n-(i+j)>=0){count++;}

    //Return the number of ways
    return count;
}

// Driver Function
int main()
{
    ll n = 3;
    cout << count_of_ways(n) << endl;
    return 0;
}

Python3

# Python program to count number of ways to break
# a number in three parts.

# Function to count number of ways
# to make the given number n
def count_of_ways(n):
    count = 0
    for i in range(n+1):
        for j in range(n+1):
            # Third number satisfying the input condition
            # will be n-(i+j) if n-(i+j)>=0
            if n-(i+j)>=0:
                count += 1

    # Return the number of ways
    return count

# Driver Function
if __name__ == '__main__':
    n = 3
    print(count_of_ways(n))

Output

10

Time Complexity : O(n²) ,because of two nested loops
Auxiliary Space: O(1) , because no extra space has been used

Efficient Approach: If we carefully observe the test cases then we realize that the number of ways to break a number n into 3 parts is equal to (n+1) * (n+2) / 2. 

C++

// C++ program to count number of ways to break
// a number in three parts.
#include <bits/stdc++.h>
#define ll long long int
using namespace std;

// Function to count number of ways
// to make the given number n
ll count_of_ways(ll n)
{
    long long int mod = 1000000007;
        int count = ((n + 1) % mod * (n + 2) % mod) / 2;
        return count;
}

// Driver Function
int main()
{
    ll n = 3;
    cout << count_of_ways(n) << endl;
    return 0;
}

Java

// Java program to count number of ways to break
// a number in three parts
import java.io.*;

class GFG {
    // Function to count number of ways
    // to make the given number n
    static long count_of_ways(long n)
    {
        long count = 0;
        count = (n + 1) * (n + 2) / 2;
        return count;
    }

    // driver program
    public static void main(String[] args)
    {
        long n = 3;
        System.out.println(count_of_ways(n));
    }
}

// Contributed by Pramod Kumar

Python3

# Python 3 program to count number of
# ways to break a number in three parts.

# Function to count number of ways
# to make the given number n
def count_of_ways(n):
    count = 0
    count = (n + 1) * (n + 2) // 2
    return count

# Driver code
n = 3
print(count_of_ways(n))

# This code is contributed by Shrikant13

C#

// C# program to count number of ways to
// break a number in three parts
using System;

class GFG {

    // Function to count number of ways
    // to make the given number n
    static long count_of_ways(long n)
    {
        long count = 0;
        count = (n + 1) * (n + 2) / 2;
        return count;
    }

    // driver program
    public static void Main()
    {
        long n = 3;
        Console.WriteLine(count_of_ways(n));
    }
}

// This code is Contributed by vt_m.

PHP

<?php
// PHP program to count number
// of ways to break a number
// in three parts.

// Function to count number of ways
// to make the given number n
function count_of_ways( $n)
{
    $count;
    $count = ($n + 1) * ($n + 2) / 2;
    return $count;
}

// Driver Code
$n = 3;
echo count_of_ways($n);

// This code is Contributed by vt_m.
?>

Javascript

<script>

// javascript program to count number of ways to
// break a number in three parts


    // Function to count number of ways
    // to make the given number n

    function count_of_ways(n)
    {
        var count = 0;
        count = (n + 1) * (n + 2) / 2;
        return count;
    }

    // driver program

        var n = 3;
        document.write(count_of_ways(n));

// This code is contributed by bunnyram19.
</script>

Output

10

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

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