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

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. 

C++

// C++ program to find the number of
// right triangles with given perimeter
#include<bits/stdc++.h>
using namespace std;

// Function to return the count
int countTriangles(int p)
{
  // making a list to store (a, b) pairs
  vector<pair<int,int>> store;

  // no triangle if p is odd
  if (p % 2 != 0)
    return 0;
  else
  {
    int count = 1;

    for(int b = 1; b < p / 2; b++)
    {
      float a = (float)p / 2.0f * ((float)((float)p -
                                           2.0 * (float)b) /
                                   ((float)p - (float)b));

      int inta = (int)(a);

      if (a == inta)
      {
        // make (a, b) pair in sorted order
        pair<int,int> ab;

        if(inta<b)
        {
          ab = {inta, b};
        }
        else
        {
          ab = {b, inta};
        }

        // check to avoid duplicates
        if(find(store.begin(), store.end(), ab) == store.end())
        {
          count += 1;

          // store the new pair
          store.push_back(ab);
        }
      }

    }
    return count;
  }
}

// Driver Code
int main()
{
  int p = 840;
  cout << "number of right triangles = " << countTriangles(p);
  return 0;
}

// This code is contributed by rutvik_56.

Java

// Java program to find the number of
// right triangles with given perimeter
import java.util.*;

class GFG{
  static class pair
  {
    int first, second;
    public pair(int first, int second)
    {
      this.first = first;
      this.second = second;
    }
  }

  // Function to return the count
  static int countTriangles(int p)
  {
    // making a list to store (a, b) pairs
    HashSet<pair> store = new HashSet<pair>();

    // no triangle if p is odd
    if (p % 2 != 0)
      return 0;
    else
    {
      int count = 1;

      for(int b = 1; b < p / 3; b++)
      {
        float a = (float)p / 2.0f * ((float)((float)p -
                                             2.0 * (float)b) /
                                     ((float)p - (float)b));

        int inta = (int)(a);

        if (a == inta)
        {
          // make (a, b) pair in sorted order
          pair ab;
          if(inta<b)
          {
            ab = new pair(inta, b);

          }
          else
          {
            ab = new pair(b, inta);

          }

          // check to astatic void duplicates
          if(!store.contains(ab) )
          {
            count += 1;

            // store the new pair
            store.add(ab);
          }
        }

      }
      return count;
    }
  }

  // Driver Code
  public static void main(String[] args)
  {
    int p = 840;
    System.out.print("number of right triangles = " +  countTriangles(p));
  }
}

// This code is contributed by Rajput-Ji

Python3

# 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)))

C#

// C# program to find the number of
// right triangles with given perimeter
using System;
using System.Collections.Generic;

public class GFG {
  public class pair {
    public int first, second;

    public pair(int first, int second) {
      this.first = first;
      this.second = second;
    }
  }

  // Function to return the count
  static int countTriangles(int p)
  {

    // making a list to store (a, b) pairs
    HashSet<pair> store = new HashSet<pair>();

    // no triangle if p is odd
    if (p % 2 != 0)
      return 0;
    else {
      int count = 1;

      for (int b = 1; b < p / 3; b++) {
        float a = (float) p / 3 * ((float) ((float) p -
                                            2 * (float) b) /
                                   ((float) p - (float) b));

        int inta = (int) (a);

        if (a == inta)
        {

          // make (a, b) pair in sorted order
          pair ab;
          if (inta < b) {
            ab = new pair(inta, b);

          } else {
            ab = new pair(b, inta);

          }

          // check to astatic void duplicates
          if (!store.Contains(ab)) {
            count += 1;

            // store the new pair
            store.Add(ab);
          }
        }

      }
      return count;
    }
  }

  // Driver Code
  public static void Main(String[] args) {
    int p = 840;
    Console.Write("number of right triangles = " + countTriangles(p));
  }
}

// This code is contributed by gauravrajput1

Javascript

<script>
// Javascript program to find the number of
// right triangles with given perimeter
     class pair {
         constructor(first , second) {
            this.first = first;
            this.second = second;
        }
    }

    // Function to return the count
    function countTriangles(p)
    {

        // making a list to store (a, b) pairs
        var store = new Set();

        // no triangle if p is odd
        if (p % 2 != 0)
            return 0;
        else {
            var count = 1;

            for (var b = 1; b < p / 3; b++) {
                var a =  p / 3 * ( ( p - 2 *  b) / ( p -  b));

                var inta = parseInt( a);

                if (a == inta) {
                    // make (a, b) pair in sorted order
                    var ab;
                    if (inta < b) {
                        ab = new pair(inta, b);

                    } else {
                        ab = new pair(b, inta);

                    }

                    // check to afunction duplicates
                    if (!store.has(ab)) {
                        count += 1;

                        // store the new pair
                        store.add(ab);
                    }
                }

            }
            return count;
        }
    }

    // Driver Code
        var p = 840;
        document.write("number of right triangles = " + countTriangles(p));

// This code is contributed by Rajput-Ji
</script>

Output:

number of right triangles = 8

Time complexity: O(P)

Space complexity: O(n) as auxiliary space is being used