Count number of triangles possible with length of sides not exceeding N

Given an integer N, the task is to find the total number of right angled triangles that can be formed such that the length of any side of the triangle is at most N.

A right-angled triangle satisfies the following condition: X2 + Y2 = Z2 where Z represents the length of the hypotenuse, and X and Y represent the lengths of the remaining two sides. 

Examples: 

Input: N = 5 
Output:
Explanation: 
The only possible combination of sides which form a right-angled triangle is {3, 4, 5}.

Input: N = 10 
Output:
Explanation: 
Possible combinations of sides which form a right-angled triangle are {3, 4, 5} and {6, 8, 10}.



Naive Approach: The idea is to generate every possible combination of triplets with integers from the range [1, N] and for each such combination, check whether it is a right-angled triangle or not. 

Below is the implementation of the above approach:

C++

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// C++ implementation of
// the above approach
 
#include<bits/stdc++.h>
using namespace std;
 
// Function to count total
// number of right angled triangle
int right_angled(int n)
{
    // Initialise count with 0
    int count = 0;
 
    // Run three nested loops and
    // check all combinations of sides
    for (int z = 1; z <= n; z++) {
        for (int y = 1; y <= z; y++) {
            for (int x = 1; x <= y; x++) {
 
                // Condition for right
                // angled triangle
                if ((x * x) + (y * y) == (z * z)) {
 
                    // Increment count
                    count++;
                }
            }
        }
    }
 
    return count;
}
 
// Driver Code
int main()
{
    // Given N
    int n = 5;
 
    // Function Call
    cout << right_angled(n);
    return 0;
}

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Java

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// Java implementation of 
// the above approach
import java.io.*;
 
class GFG{
     
// Function to count total
// number of right angled triangle
static int right_angled(int n)
{
     
    // Initialise count with 0
    int count = 0;
     
    // Run three nested loops and
    // check all combinations of sides
    for(int z = 1; z <= n; z++)
    {
        for(int y = 1; y <= z; y++)
        {
            for(int x = 1; x <= y; x++)
            {
                 
                // Condition for right
                // angled triangle
                if ((x * x) + (y * y) == (z * z))
                {
                     
                    // Increment count
                    count++;
                }
            }
        }
    }
    return count;
}
 
// Driver code
public static void main (String[] args)
{
     
    // Given N
    int n = 5;
 
    // Function call
    System.out.println(right_angled(n));
}
}
 
// This code is contributed by code_hunt

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Python3

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# Python implementation of 
# the above approach
 
# Function to count total
# number of right angled triangle
def right_angled(n):
     
    # Initialise count with 0
    count = 0
 
    # Run three nested loops and
    # check all combinations of sides
    for z in range(1, n + 1):
        for y in range(1, z + 1):
            for x in range(1, y + 1):
 
                # Condition for right
                # angled triangle
                if ((x * x) + (y * y) == (z * z)):
 
                    # Increment count
                    count += 1
                 
    return count
 
# Driver Code
 
# Given N
n = 5
 
# Function call
print(right_angled(n))
 
# This code is contributed by code_hunt

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C#

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// C# implementation of
// the above approach
using System;
 
class GFG{
     
// Function to count total
// number of right angled triangle
static int right_angled(int n)
{
     
    // Initialise count with 0
    int count = 0;
     
    // Run three nested loops and
    // check all combinations of sides
    for(int z = 1; z <= n; z++)
    {
        for(int y = 1; y <= z; y++)
        {
            for(int x = 1; x <= y; x++)
            {
                 
                // Condition for right
                // angled triangle
                if ((x * x) + (y * y) == (z * z))
                {
 
                    // Increment count
                    count++;
                }
            }
        }
    }
    return count;
}
     
// Driver Code
public static void Main(string[] args)
{
     
    // Given N
    int n = 5;
 
    // Function call
    Console.Write(right_angled(n));
}
}
 
// This code is contributed by rutvik_56

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Output: 

1



 

Time complexity: O(N3
Auxiliary Space: O(1)

Efficient Approach: The above approach can be optimized based on the idea that the third side of the triangle can be found out, if the two sides of the triangles are known. Follow the steps below to solve the problem: 

  1. Iterate up to N and generate pairs of possible length of two sides and find the third side using the relation x2 + y2 = z2
  2. If sqrt(x2+y2) is found to be an integer, store the three concerned integers in a Set in sorted order, as they can form a right angled triangle.
  3. Print the final size of the set as the required count.

Below is the implementation of the above approach: 

C++

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// C++ implementation of the
// above approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to count total
// number of right angled triangle
int right_angled(int n)
{
    // Consider a set to store
    // the three sides
    set<pair<int, pair<int, int> > > s;
 
    // Find possible third side
    for (int x = 1; x <= n; x++) {
        for (int y = 1; y <= n; y++) {
 
            // Condition for a right
            // angled triangle
            if (x * x + y * y <= n * n) {
                int z = sqrt(x * x + y * y);
 
                // Check if the third side
                // is an integer
                if (z * z != (x * x + y * y))
                    continue;
 
                vector<int> v;
 
                // Push the three sides
                v.push_back(x);
                v.push_back(y);
                v.push_back(sqrt(x * x + y * y));
 
                sort(v.begin(), v.end());
 
                // Insert the three sides in
                // the set to find unique triangles
                s.insert({ v[0], { v[1], v[2] } });
            }
            else
                break;
        }
    }
 
    // return the size of set
    return s.size();
}
 
// Driver code
int main()
{
    // Given N
    int n = 5;
 
    // Function Call
    cout << right_angled(n);
    return 0;
}

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Java

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// Java implementation of the
// above approach
import java.util.*;
 
class Pair<F, S>
{
     
    // First member of pair
    private F first;
     
    // Second member of pair
    private S second;
 
    public Pair(F first, S second)
    {
        this.first = first;
        this.second = second;
    }
}
 
class GFG{
 
// Function to count total
// number of right angled triangle    
public static int right_angled(int n)
{
     
    // Consider a set to store
    // the three sides
    Set<Pair<Integer,
        Pair<Integer,
             Integer>>> s = new HashSet<Pair<Integer,
                                        Pair<Integer,
                                             Integer>>>();
  
    // Find possible third side
    for(int x = 1; x <= n; x++)
    {
        for(int y = 1; y <= n; y++)
        {
             
            // Condition for a right
            // angled triangle
            if (x * x + y * y <= n * n)
            {
                int z = (int)Math.sqrt(x * x + y * y);
  
                // Check if the third side
                // is an integer
                if (z * z != (x * x + y * y))
                    continue;
  
                Vector<Integer> v = new Vector<Integer>();
                 
                // Push the three sides
                v.add(x);
                v.add(y);
                v.add((int)Math.sqrt(x * x + y * y));
  
                Collections.sort(v);
  
                // Add the three sides in
                // the set to find unique triangles
                s.add(new Pair<Integer,
                          Pair<Integer,
                               Integer>>(v.get(0),
                      new Pair<Integer,
                               Integer>(v.get(1),
                                        v.get(2))));
            }
            else
                break;
        }
    }
     
    // Return the size of set
    return s.size() - 1;
}
  
// Driver code
public static void main(String[] args)
{
     
    // Given N
    int n = 5;
  
    // Function call
    System.out.println(right_angled(n));
}
}
 
// This code is contributed by grand_master

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Output: 

1



 

Time complexity: O(N2
Auxiliary Space: O(1)

 

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