# 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++ implementation of ` `// the above approach`   `#include` `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 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`

 `# 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# 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`

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++ implementation of the ` `// above approach ` `#include ` `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 > > 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 implementation of the ` `// above approach ` `import` `java.util.*;`   `class` `Pair ` `{` `    `  `    ``// 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>> s = ``new` `HashSet>>();` ` `  `    ``// 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 v = ``new` `Vector();` `                `  `                ``// 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>(v.get(``0``),` `                      ``new` `Pair(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`

Output:
```1

```

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

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.