# Count number of right triangles possible with a given perimeter

Given a perimeter P, the task is to find the number of right triangles possible with perimeter equal to p.

Examples:

```Input: P = 12
Output: number of right triangles = 1
The only right angle possible is with sides
hypotenuse = 5, perpendicular = 4 and base = 3.

Input: p = 840
Output: number of right triangles = 8
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

So the aim is to find the number of solutions which satisfy equations a + b + c = p and a2 + b2 = c2.

A naive approach is to run two loops for a(1 to p/2) and b(a+1 to p/3) then make c=p-a-b and increase count by one if . This will take time.

An efficient approach can be found by little algebraic manipulation :

Since a + c > b or, p – b > b or, b < p/2. Thus iterating b from 1 to p/2, calculating a and storing only the whole number a would give all solutions for a given p. There are no right triangles are possible for odd p as right angle triangles follow the Pythagoras theorem. Use a list of pairs to store the values of a and band return the count at the end.

Below is the implementation of the above approach.

 `# python program to find the number of ` `# right triangles with given perimeter ` ` `  `# Function to return the count  ` `def` `countTriangles(p): ` `     `  `    ``# making a list to store (a, b) pairs ` `    ``store ``=``[] ` ` `  `    ``# no triangle if p is odd ` `    ``if` `p ``%` `2` `!``=` `0` `: ``return` `0` `    ``else` `: ` `        ``count ``=` `0` `        ``for` `b ``in` `range``(``1``, p ``/``/` `2``): ` ` `  `            ``a ``=` `p ``/` `2` `*` `((p ``-` `2` `*` `b) ``/` `(p ``-` `b)) ` `            ``inta ``=` `int``(a) ` `            ``if` `(a ``=``=` `inta ): ` ` `  `                ``# make (a, b) pair in sorted order  ` `                ``ab ``=` `tuple``(``sorted``((inta, b))) ` ` `  `                ``# check to avoid duplicates ` `                ``if` `ab ``not` `in` `store : ` `                    ``count ``+``=` `1` `                    ``# store the new pair  ` `                    ``store.append(ab) ` `        ``return` `count ` ` `  `# Driver Code ` `p ``=` `840` `print``(``"number of right triangles = "``+``str``(countTriangles(p))) `

Output:
```number of right triangles = 8
```

Time complexity: O(P)

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.

Improved By : siddhanthapliyal

Article Tags :