Number of non-negative integral solutions of a + b + c = n
Given a number n, find the number of ways in which we can add 3 non-negative integers so that their sum is n.
Examples :
Input : n = 1
Output : 3
There are three ways to get sum 1.
(1, 0, 0), (0, 1, 0) and (0, 0, 1)
Input : n = 2
Output : 6
There are six ways to get sum 2.
(2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 1, 0)
(1, 0, 1) and (0, 1, 1)
Input : n = 3
Output : 10
There are ten ways to get sum 3.
(3, 0, 0), (0, 3, 0), (0, 0, 3), (1, 2, 0)
(1, 0, 2), (0, 1, 2), (2, 1, 0), (2, 0, 1),
(0, 2, 1) and (1, 1, 1)
Method 1 [ Brute Force : O(n3) ]
A simple solution is to consider all triplets using three loops. For every triplet, check if its sum is equal to n. If the sum is n, increment the count of solutions.
Below is the implementation.
C++
#include<bits/stdc++.h>
using namespace std;
int countIntegralSolutions( int n)
{
int result = 0;
for ( int i=0; i<=n; i++)
for ( int j=0; j<=n-i; j++)
for ( int k=0; k<=(n-i-j); k++)
if (i + j + k == n)
result++;
return result;
}
int main()
{
int n = 3;
cout << countIntegralSolutions(n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int countIntegralSolutions( int n)
{
int result = 0 ;
for ( int i = 0 ; i <= n; i++)
for ( int j = 0 ; j <= n - i; j++)
for ( int k = 0 ; k <= (n - i - j); k++)
if (i + j + k == n)
result++;
return result;
}
public static void main (String[] args)
{
int n = 3 ;
System.out.println( countIntegralSolutions(n));
}
}
|
Python3
def countIntegralSolutions (n):
result = 0
for i in range (n + 1 ):
for j in range (n + 1 ):
for k in range (n + 1 ):
if i + j + k = = n:
result + = 1
return result
n = 3
print (countIntegralSolutions(n))
|
C#
using System;
class GFG {
static int countIntegralSolutions( int n)
{
int result = 0;
for ( int i = 0; i <= n; i++)
for ( int j = 0; j <= n - i; j++)
for ( int k = 0; k <= (n - i - j); k++)
if (i + j + k == n)
result++;
return result;
}
public static void Main ()
{
int n = 3;
Console.Write(countIntegralSolutions(n));
}
}
|
PHP
<?php
function countIntegralSolutions( $n )
{
$result = 0;
for ( $i = 0; $i <= $n ; $i ++)
for ( $j = 0; $j <= $n - $i ; $j ++)
for ( $k = 0; $k <= ( $n - $i - $j ); $k ++)
if ( $i + $j + $k == $n )
$result ++;
return $result ;
}
$n = 3;
echo countIntegralSolutions( $n );
?>
|
Javascript
<script>
function countIntegralSolutions(n)
{
let result = 0;
for (let i = 0; i <= n; i++)
for (let j = 0; j <= n - i; j++)
for (let k = 0; k <= (n - i - j); k++)
if (i + j + k == n)
result++;
return result;
}
let n = 3;
document.write(countIntegralSolutions(n));
</script>
|
Output:
10
Time complexity: O(n3)
Auxiliary Space: O(1)
Method 2 [ Direct Formula : O(1) ]
If we take a closer look at the pattern, we can find that the count of solutions is ((n+1) * (n+2)) / 2. The problem is equivalent to distributing n identical balls in three boxes and the solution is n+2C2. In general, if there are m variables (or boxes) and n balls , the formula becomes n+m-1Cm-1.
C++
#include<bits/stdc++.h>
using namespace std;
int countIntegralSolutions( int n)
{
return ((n+1)*(n+2))/2;
}
int main()
{
int n = 3;
cout << countIntegralSolutions(n);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int countIntegralSolutions( int n)
{
return ((n + 1 ) * (n + 2 )) / 2 ;
}
public static void main (String[] args)
{
int n = 3 ;
System.out.println ( countIntegralSolutions(n));
}
}
|
Python3
def countIntegralSolutions (n):
return int (((n + 1 ) * (n + 2 )) / 2 )
n = 3
print (countIntegralSolutions(n))
|
C#
using System;
class GFG {
static int countIntegralSolutions( int n)
{
return ((n + 1) * (n + 2)) / 2;
}
public static void Main (String[] args)
{
int n = 3;
Console.Write ( countIntegralSolutions(n));
}
}
|
PHP
<?php
function countIntegralSolutions( $n )
{
return (( $n + 1) * ( $n + 2)) / 2;
}
$n = 3;
echo countIntegralSolutions( $n );
?>
|
Javascript
<script>
function countIntegralSolutions(n)
{
return Math.floor(((n+1)*(n+2))/2);
}
let n = 3;
document.write(countIntegralSolutions(n));
</script>
|
Output :
10
Time complexity: O(1)
Auxiliary Space: O(1)
Last Updated :
24 Jun, 2022
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