Count non-negative triplets with sum equal to N

Last Updated : 29 Mar, 2023

Given an integer N. The task is to find the number of different ordered triplets(a, b, c) of non-negative integers such that a + b + c = N .
Examples:

Input : N = 2
Output : 6
Triplets are : (0, 0, 2), (1, 0, 1), (0, 1, 1), (2, 0, 0), (0, 2, 0), (1, 1, 0)
Input : N = 50
Output : 1326

Naive Approach:

We can use three nested loops to generate all possible ordered triplets of non-negative integers such that their sum is equal to the given number N.

Implementation of the above approach:

C++

#include <iostream>
using namespace std;

// Function to calculate number of triplets
int countTriplets(int N)
{
    // Initializing result
    int res = 0;

    // Finding all possible combinations
    // using nested loops
    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)
                    res++;
            }
        }
    }

    // Returning result
    return res;
}

// Driver code
int main()
{
    int N = 50;

    // Function call
    cout << countTriplets(N);

    return 0;
}

Java

import java.util.Scanner;

public class Main {

    // Function to calculate number of triplets
    public static int countTriplets(int N) {
        // Initializing result
        int res = 0;

        // Finding all possible combinations
        // using nested loops
        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)
                        res++;
                }
            }
        }

        // Returning result
        return res;
    }

    // Driver code
    public static void main(String[] args) {
        int N = 50;

        // Function call
        System.out.println(countTriplets(N));
    }
}
// This code is contributed by Prajwal Kandekar

Python3

def countTriplets(N):
  
    # Initializing result
    res = 0

    # Finding all possible combinations
    # using nested loops
    for i in range(N+1):
        for j in range(N+1):
            for k in range(N+1):
                if i + j + k == N:
                    res += 1

    # Returning result
    return res

# Driver code
N = 50

# Function call
print(countTriplets(N))

# This code is contributed by Prajwal Kandekar

using System;

public class GFG{

   // Function to calculate number of triplets
public static int CountTriplets(int N) {
    // Initializing result
    int res = 0;

    // Finding all possible combinations
    // using nested loops
    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)
                    res++;
            }
        }
    }

    // Returning result
    return res;
}

// Driver code
public static void Main(string[] args) {
    int N = 50;

    // Function call
    Console.WriteLine(CountTriplets(N));
}
    }

JavaScript

// Function to calculate number of triplets
function countTriplets(N) {
    // Initializing result
    let res = 0;

    // Finding all possible combinations
    // using nested loops
    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)
                    res++;
            }
        }
    }

    // Returning result
    return res;
}

// Driver code
let N = 50;

// Function call
console.log(countTriplets(N));

Output

Time Complexity: O(N^3)

Auxiliary Space: O(1)

Efficient Approach :
First, it is easy to see that for each non-negative integer N, the equation a + b = N can be satisfied by (N+1) different ordered pairs of (a, b). Now we can assign c values from 0 to N then the ordered pairs for a+b can be found. It will form a series of N+1 natural numbers and its sum will give the count of triplets.
Below is the implementation of the above approach :

C++

// CPP program to find triplets count
#include <bits/stdc++.h>
using namespace std;

// Function to find triplets count
int triplets(int N)
{
    // Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) / 2;
}

// Driver code
int main()
{
    int N = 50;
    
    // Function call
    cout << triplets(N);
    
    return 0;
}

Java

// Java program to find triplets count
class GFG 
{
    
// Function to find triplets count
static int triplets(int N)
{
    // Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) / 2;
}

// Driver code
public static void main(String[] args)
{
    int N = 50;

    System.out.println(triplets(N));
}
}

// This code is contributed 
// by PrinciRaj1992

Python3

# Python3 program to find triplets count

# Function to find triplets count
def triplets(N):

    # Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) // 2;

# Driver code
N = 50;
    
# Function call
print(triplets(N))

# This code is contributed by nidhi

// C# program to find triplets count
using System;

class GFG 
{
    
// Function to find triplets count
static int triplets(int N)
{
    // Sum of first n+1 natural numbers
    return ((N + 1) * (N + 2)) / 2;
}

// Driver code
public static void Main()
{
    int N = 50;

    Console.WriteLine(triplets(N));
}
}

// This code is contributed 
// by anuj_67..

JavaScript

<script>

    // Javascript program to find triplets count    
    
    // Function to find triplets count 
    function triplets(N) 
    { 
        // Sum of first n+1 natural numbers 
        return ((N + 1) * (N + 2)) / 2; 
    } 
    
    let N = 50;
    document.write(triplets(N));
    
</script>

Output

Time Complexity : O(1)

Auxiliary Space: O(1)

Count non-negative triplets with sum equal to N

isa_aanchal

Improve

Article Tags :

Practice Tags :

Count non-negative triplets with sum equal to N

Similar Reads

Thank You!

What kind of Experience do you want to share?