Number of non-negative integral solutions of a + b + c = n

Given a number n, find number of ways we can add 3 non-negative integers so that their sum is n.

Examples :

Input : n = 1
Output : 3
There are four 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 10.
(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 solutions is to consider all triplets using three loops. For every triplet, check if its sum is equal to n. If sum is n, increment count of solutions.

Below is the implementation.

C++

// A naive C++ solution to count solutions of
// a + b + c = n
#include<bits/stdc++.h>
using namespace std;
  
// Returns count of solutions of a + b + c = n
int countIntegralSolutions(int n)
{
    // Initialize result
    int result = 0;
  
    // Consider all triplets and increment
    // result whenever sum of a triplet is n.
    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;
}
  
// Driver code
int main()
{
    int n = 3;
    cout <<  countIntegralSolutions(n);
    return 0;
}

Java

// A naive Java solution to count
// solutions of a + b + c = n
import java.io.*;
  
class GFG 
{
    // Returns count of solutions of a + b + c = n
    static int countIntegralSolutions(int n)
    {
        // Initialize result
        int result = 0;
      
        // Consider all triplets and increment
        // result whenever sum of a triplet is n.
        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;
    }
  
    // Driver code
    public static void main (String[] args) 
    {
        int n = 3;
        System.out.println( countIntegralSolutions(n));
      
    }
}
  
// This article is contributed by vt_m

Python3

# Python3 code to count 
# solutions of a + b + c = n
  
# Returns count of solutions
# of a + b + c = n
def countIntegralSolutions (n):
  
    # Initialize result
    result = 0
      
    # Consider all triplets and 
    # increment result whenever 
    # sum of a triplet is n.
    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
      
# Driver code
n = 3
print(countIntegralSolutions(n))
  
# This code is contributed by "Sharad_Bhardwaj".

C#

// A naive C# solution to count
// solutions of a + b + c = n
using System;
  
class GFG {
      
    // Returns count of solutions
    // of a + b + c = n
    static int countIntegralSolutions(int n)
    {
          
        // Initialize result
        int result = 0;
      
        // Consider all triplets and increment
        // result whenever sum of a triplet is n.
        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;
    }
  
    // Driver code
    public static void Main () 
    {
        int n = 3;
        Console.Write(countIntegralSolutions(n));
      
    }
}
  
// This article is contributed by Smitha.

PHP

<?php
// A naive PHP solution
// to count solutions of
// a + b + c = n
  
// Returns count of 
// solutions of a + b + c = n
function countIntegralSolutions( $n)
{
      
    // Initialize result
    $result = 0;
  
    // Consider all triplets and increment
    // result whenever sum of a triplet is n.
    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;
}
  
    // Driver Code
    $n = 3;
    echo countIntegralSolutions($n);
  
// This code is contributed by anuj_67.
?>


Output :

10

 

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 + 1 identical balls (for 0 to n) in three boxes and the solution is n+2C2. In general, if there are m variables (or boxes) and n possible values, the formula becomes n+m-1Cm-1.

C++

// A naive C++ solution to count solutions of
// a + b + c = n
#include<bits/stdc++.h>
using namespace std;
  
// Returns count of solutions of a + b + c = n
int countIntegralSolutions(int n)
{
    return ((n+1)*(n+2))/2;
}
  
// Driver code
int main()
{
    int n = 3;
    cout <<  countIntegralSolutions(n);
    return 0;
}

Java

// Java solution to count 
// solutions of a + b + c = n
import java.io.*;
  
class GFG 
{
    // Returns count of solutions 
    // of a + b + c = n
    static int countIntegralSolutions(int n)
    {
    return ((n + 1) * (n + 2)) / 2;
          
    }
      
    // Driver code
    public static void main (String[] args) 
    {
        int n = 3;
        System.out.println ( countIntegralSolutions(n));
          
    }
}
// This article is contributed by vt_m

Python3

# Python3 solution to count 
# solutions of a + b + c = n
  
# Returns count of solutions
# of a + b + c = n
def countIntegralSolutions (n):
    return int(((n + 1) * (n + 2)) / 2)
      
# Driver code
n = 3
print(countIntegralSolutions(n))
  
# This code is contributed by "Sharad_Bhardwaj".

C#

// C# solution to count 
// solutions of a + b + c = n
using System;
  
class GFG {
      
    // Returns count of solutions 
    // of a + b + c = n
    static int countIntegralSolutions(int n)
    {
        return ((n + 1) * (n + 2)) / 2;
    }
      
    // Driver code
    public static void Main (String[] args) 
    {
        int n = 3;
        Console.Write ( countIntegralSolutions(n));
          
    }
}
  
// This code is contributed by parashar.

PHP

<?php
// A naive PHP solution 
// to count solutions of
// a + b + c = n
  
// Returns count of solutions
// of a + b + c = n
function countIntegralSolutions($n)
{
    return (($n + 1) * ($n + 2)) / 2;
}
  
    // Driver Code
    $n = 3;
    echo countIntegralSolutions($n);
  
// This code is contributed by anuj_67.
?>


Output :

10

This article is contributed by Shivam Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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