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Count of ways to split N into Triplets forming a Triangle

  • Last Updated : 11 Jun, 2021

Given an integer N, the task is to find the number of ways to split N into ordered triplets which can together form a triangle.

Examples:

Input: N = 15 
Output: Total number of triangles possible are 28

Input: N = 9 
Output: Total number of triangles possible is 10 
 

Approach: The following observation needs to be made in order to solve the problem: 



If N is split into 3 integers a, b and c, then the following conditions need to be satisfied for a, b and c to form a triangle: 

  • a + b > c
     
  • a + c > b
     
  • b + c > a

Therefore, iterate over the range [1, N] using nested loops to generate triplets, and for each triplet check if it forms a triangle or not. 

Below is the implementation of the above approach:

C++




// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the
// required number of ways
int Numberofways(int n)
{
    int count = 0;
 
    for (int a = 1; a < n; a++) {
 
        for (int b = 1; b < n; b++) {
 
            int c = n - (a + b);
 
            // Check if a, b and c can
            // form a triangle
            if (a + b > c && a + c > b
                && b + c > a) {
                count++;
            }
        }
    }
 
    // Return number of ways
    return count;
}
 
// Driver Code
int main()
{
    int n = 15;
 
    cout << Numberofways(n) << endl;
 
    return 0;
}

Java




// Java Program to implement
// the above approach
import java.io.*;
 
class GFG {
 
    // Function to return the
    // required number of ways
    static int Numberofways(int n)
    {
        int count = 0;
 
        for (int a = 1; a < n; a++) {
 
            for (int b = 0; b < n; b++) {
 
                int c = n - (a + b);
 
                // Check if a, b, c can
                // form a triangle
                if (a + b > c && a + c > b
                    && b + c > a) {
                    count++;
                }
            }
        }
 
        return count;
    }
 
    // Driver Code
    public static void main(String[] args)
    {
        int n = 15;
 
        System.out.println(Numberofways(n));
    }
}

Python3




# Python Program to implement
# the above approach
 
# Function to return the
# required number of ways
def Numberofways(n):
    count = 0
    for a in range(1, n):
        for b in range(1, n):
 
            c = n - (a + b)
 
            # Check if a, b, c can form a triangle
            if(a < b + c and b < a + c and c < a + b):
                count += 1
 
    return count
 
 
# Driver code
n = 15
print(Numberofways(n))

C#




// C# Program to implement
// the above approach
 
using System;
 
class GFG {
 
    // Function to return the
    // required number of ways
    static int Numberofways(int n)
    {
        int count = 0;
        for (int a = 1; a < n; a++) {
            for (int b = 1; b < n; b++) {
                int c = n - (a + b);
 
                // Check if a, b, c can form
                // a triangle or not
                if (a + b > c && a + c > b
                    && b + c > a) {
                    count++;
                }
            }
        }
 
        // Return number of ways
        return count;
    }
 
    // Driver Code
    static public void Main()
    {
        int n = 15;
 
        Console.WriteLine(Numberofways(n));
    }
}

Javascript




<script>
// Javascript Program to implement
// the above approach
 
// Function to return the
// required number of ways
function Numberofways(n)
{
    var count = 0;
 
    for (var a = 1; a < n; a++)
    {
        for (var b = 1; b < n; b++)
        {
            var c = n - (a + b);
 
            // Check if a, b and c can
            // form a triangle
            if (a + b > c && a + c > b
                && b + c > a)
            {
                count++;
            }
        }
    }
 
    // Return number of ways
    return count;
}
 
// Driver Code
var n = 15;
document.write( Numberofways(n));
 
// This code is contributed by noob2000.
</script>
Output: 
28

 

Time Complexity: O(N2
Auxiliary Space: O(N)

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