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: 1
Explanation:
The only possible combination of sides which form a right-angled triangle is {3, 4, 5}.Input: N = 10
Output: 2
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++
// 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; } |
Java
// 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 |
Python3
# 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 |
C#
// 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 |
Javascript
<script> // javascript implementation of // the above approach // Function to count total // number of right angled triangle function right_angled(n) { // Initialise count with 0 var count = 0; // Run three nested loops and // check all combinations of sides for (z = 1; z <= n; z++) { for (y = 1; y <= z; y++) { for (x = 1; x <= y; x++) { // Condition for right // angled triangle if ((x * x) + (y * y) == (z * z)) { // Increment count count++; } } } } return count; } // Driver code //Given N var n = 5; // Function call document.write(right_angled(n)); // This code is contributed by Amit Katiyar </script> |
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:
- Iterate up to N and generate pairs of possible length of two sides and find the third side using the relation x2 + y2 = z2
- 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.
- Print the final size of the set as the required count.
Below is the implementation of the above approach:
C++
// 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; } |
Java
// 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 |
Python3
# Python implementation of the # above approach import math # Function to count total # number of right angled triangle def right_angled(n): # Consider a set to store # the three sides s = {} # Find possible third side for x in range ( 1 ,n + 1 ): for y in range ( 1 ,n + 1 ): # Condition for a right # angled triangle if (x * x + y * y< = n * n): z = int (math.sqrt(x * x + y * y)) # Check if the third side # is an integer if (z * z! = (x * x + y * y)): continue v = [] # Push the three sides v.append(x) v.append(y) v.append( int (math.sqrt(x * x + y * y))) v.sort() # Insert the three sides in # the set to find unique triangles s[v[ 0 ]] = [v[ 1 ],v[ 2 ]] else : break # return the size of set return len (s) # Driver code # Given N n = 5 # Function Call print (right_angled(n)) # This code is contributed by Aman Kumar. |
Javascript
// JavaScript implementation of the // above approach // Function to count total // number of right angled triangle function right_angled(n) { // Consider a set to store // the three sides let s = new Set(); // Find possible third side for (let x = 1; x <= n; x++) { for (let y = 1; y <= n; y++) { // Condition for a right // angled triangle if (x * x + y * y <= n * n) { let z = Math.floor(Math.sqrt(x * x + y * y)); // Check if the third side // is an integer if (z * z != (x * x + y * y)) continue ; let v = new Array(); // Push the three sides v.push(x); v.push(y); v.push(Math.floor(Math.sqrt(x * x + y * y))); v.sort(); // Insert the three sides in // the set to find unique triangles s.add([v[0],[v[1], v[2]]].join()); } else break ; } } // return the size of set return s.size; } // Driver code // Given N let n = 5; // Function Call console.log(right_angled(n)); // The code is contributed by Gautam goel (gautamgoel962) |
C#
// C# implementation of the // above approach using System; using System.Collections.Generic; 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 HashSet<Pair< int , Pair< int , int > > > s = new HashSet<Pair< int , Pair< int , int > > >(); // 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 ; List< int > v = new List< int >(); // Push the three sides v.Add(x); v.Add(y); v.Add(( int )Math.Sqrt(x * x + y * y)); v.Sort(); // Add the three sides in // the set to find unique triangles s.Add( new Pair< int , Pair< int , int > >( v[0], new Pair< int , int >(v[1], v[2]))); } else break ; } } // Return the size of set return s.Count - 1; } // Driver code public static void Main( string [] args) { // Given N int n = 5; // Function call Console.WriteLine(right_angled(n)); } } // This code is contributed by phasing17 |
1
Time complexity: O(N2*log(N)) as using sqrt inside inner for loop
Auxiliary Space: O(N) since using auxiliary space for set
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