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++
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
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;
}
int main()
{
ll n = 3;
cout << count_of_ways(n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
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;
}
public static void main(String[] args)
{
long n = 3 ;
System.out.println(count_of_ways(n));
}
}
|
Python3
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
if __name__ = = '__main__' :
n = 3
print (count_of_ways(n))
|
C#
using System;
class GFG {
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;
}
public static void Main()
{
long n = 3;
Console.WriteLine(count_of_ways(n));
}
}
|
PHP
<?php
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 ;
}
$n = 3;
echo count_of_ways( $n );
?>
|
Javascript
<script>
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;
}
let n = 3;
document.write(count_of_ways(n) + "<br>" );
</script>
|
Time Complexity : O(n3)
Auxiliary Space: O(1)
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++
#include <bits/stdc++.h>
#define ll long long int
using namespace std;
ll count_of_ways(ll n)
{
long long int mod = 1000000007;
int count = ((n + 1) % mod * (n + 2) % mod) / 2;
return count;
}
int main()
{
ll n = 3;
cout << count_of_ways(n) << endl;
return 0;
}
|
Java
import java.io.*;
class GFG {
static long count_of_ways( long n)
{
long count = 0 ;
count = (n + 1 ) * (n + 2 ) / 2 ;
return count;
}
public static void main(String[] args)
{
long n = 3 ;
System.out.println(count_of_ways(n));
}
}
|
Python3
def count_of_ways(n):
count = 0
count = (n + 1 ) * (n + 2 ) / / 2
return count
n = 3
print (count_of_ways(n))
|
C#
using System;
class GFG {
static long count_of_ways( long n)
{
long count = 0;
count = (n + 1) * (n + 2) / 2;
return count;
}
public static void Main()
{
long n = 3;
Console.WriteLine(count_of_ways(n));
}
}
|
PHP
<?php
function count_of_ways( $n )
{
$count ;
$count = ( $n + 1) * ( $n + 2) / 2;
return $count ;
}
$n = 3;
echo count_of_ways( $n );
?>
|
Javascript
<script>
function count_of_ways(n)
{
var count = 0;
count = (n + 1) * (n + 2) / 2;
return count;
}
var n = 3;
document.write(count_of_ways(n));
</script>
|
Time Complexity: O(1)
Auxiliary Space: O(1)
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