Count of ways to split N into Triplets forming a Triangle Last Updated : 11 Jun, 2021 Comments Improve Suggest changes Like Article Like Report 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) Comment More infoAdvertise with us Next Article Count of ways to split N into Triplets forming a Triangle R rubaimandal3 Follow Improve Article Tags : Misc Greedy Searching Mathematical Computer Science Fundamentals DSA triangle +3 More Practice Tags : GreedyMathematicalMiscSearching Similar Reads Count Non-Path Triplets in a Tree Given a tree with n vertices numbered from 1 to n. 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