Saint-Exupery Numbers

Saint-Exupery Number a number N such that N is the product of the three sides of Pythagorean triangle.
Some of the Saint-Exupery numbers are: 

60, 480, 780, 1620, 2040, 3840, 4200, 6240, 7500, 12180…. 

Check if N is a Saint Exupery number

Given a number N, the task is to check if N is a Saint-Exupery Number or not. If N is an Saint-Exupery Number then print “Yes” else print “No”.
Examples:

Input: N = 60 
Output: Yes 
Explanation:
60 = 3 * 4 * 5 and 3^2 + 4^2 = 5^2.
Input: N = 120 
Output: No 

Naive Approach: A simple solution is to run three nested loops to generate all possible triplets and for every triplet, check if it is a Pythagorean Triplet and has given product. Time complexity of this solution is O(n³).
Efficient Approach: The idea is to run two loops, where first loop runs from i = 1 to n/3, second loop runs from j = i+1 to n/2. In second loop, we check if (n / i / j) is equal to i * i + j * j.
Below is the implementation of the above approach:

C++

// C++ implementation to check if N
// is a Saint-Exupery number

#include <bits/stdc++.h>
using namespace std;

// Function to check if a number is
//  a Saint-Exupery number
bool isSaintExuperyNum(int n)
{
    // Considering triplets in
    // sorted order. The value
    // of first element in sorted
    // triplet can be at-most n/3.
    for (int i = 1; i <= n / 3; i++) {

        // The value of second
        // element must be less
        // than equal to n/2
        for (int j = i + 1; j <= n / 2; j++) {
            int k = n / i / j;
            if (i * i + j * j == k * k) {
                if (i * j * k == n)
                    return true;
                ;
            }
        }
    }

    return false;
}

// Driver Code
int main()
{
    // Given Number N
    int N = 60;

    // Function Call
    if (isSaintExuperyNum(N))
        cout << "Yes";
    else
        cout << "No";
    return 0;
}

Java

// Java program for above approach
class GFG{

// Function to check if a number is
// a Saint-Exupery number
static boolean isSaintExuperyNum(int n)
{
    // Considering triplets in
    // sorted order. The value
    // of first element in sorted
    // triplet can be at-most n/3.
    for (int i = 1; i <= n / 3; i++)
    {

        // The value of second
        // element must be less
        // than equal to n/2
        for (int j = i + 1; j <= n / 2; j++)
        {
            int k = n / i / j;
            if (i * i + j * j == k * k)
            {
                if (i * j * k == n)
                    return true;
            }
        }
    }

    return false;
}

// Driver Code
public static void main(String[] args)
{
    // Given Number N
    int N = 60;

    // Function Call
    if (isSaintExuperyNum(N))
        System.out.print("Yes");
    else
        System.out.print("No");
}
}

// This code is contributed by Shubham Prakash

Python3

# Python3 implementation to check if N
# is a Saint-Exupery number

# Function to check if a number is
# a Saint-Exupery number
def isSaintExuperyNum(n):

    # Considering triplets in
    # sorted order. The value
    # of first element in sorted
    # triplet can be at-most n/3.
    for i in range(1, (n // 3) + 1):

        # The value of second
        # element must be less
        # than equal to n/2
        for j in range(i + 1, (n // 2) + 1):
            k = n / i / j
            if i * i + j * j == k * k:
                if i * j * k == n:
                    return True

    return False

# Driver Code

# Given Number N
N = 60

# Function Call
if isSaintExuperyNum(N):
    print("Yes")
else:
    print("No")

# This code is contributed by
# divyamohan123

C#

// C# program for above approach
using System;
class GFG{

// Function to check if a number is
// a Saint-Exupery number
static bool isSaintExuperyNum(int n)
{
    // Considering triplets in
    // sorted order. The value
    // of first element in sorted
    // triplet can be at-most n/3.
    for (int i = 1; i <= n / 3; i++)
    {

        // The value of second
        // element must be less
        // than equal to n/2
        for (int j = i + 1; j <= n / 2; j++)
        {
            int k = n / i / j;
            if (i * i + j * j == k * k)
            {
                if (i * j * k == n)
                    return true;
            }
        }
    }

    return false;
}

// Driver Code
public static void Main()
{
    // Given Number N
    int N = 60;

    // Function Call
    if (isSaintExuperyNum(N))
        Console.Write("Yes");
    else
        Console.Write("No");
}
}

// This code is contributed by Code_Mech

Javascript

<script>

// Javascript program for above approach


    // Function to check if a number is
    // a Saint-Exupery number
    function isSaintExuperyNum( n) {
        // Considering triplets in
        // sorted order. The value
        // of first element in sorted
        // triplet can be at-most n/3.
        for ( i = 1; i <= n / 3; i++) {

            // The value of second
            // element must be less
            // than equal to n/2
            for ( j = i + 1; j <= n / 2; j++) {
                let k = n / i / j;
                if (i * i + j * j == k * k) {
                    if (i * j * k == n)
                        return true;
                }
            }
        }

        return false;
    }

    // Driver Code

        // Given Number N
        let N = 60;

        // Function Call
        if (isSaintExuperyNum(N))
            document.write("Yes");
        else
            document.write("No");

// This code contributed by gauravrajput1

</script>

Output:

Yes

Time Complexity: O(n^2)
Reference: http://www.numbersaplenty.com/set/Saint-Exupery_number/