Idoneal Numbers

Idoneal Number is a number N if and only if it cannot be written as ab + bc + ca for a > b > c > 0.
Few Idoneal numbers are: 

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 15, 16, 18, 21, 22………… 

Check if a number is an Idoneal number

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

Input: N = 10 
Output: Yes

Input: N = 11 
Output: No 

Approach The idea is to run three nested loops from ‘1’ to ‘N’ and check whether ab + bc + ca equal ‘N’ or not. Return false if ab + bc + ca equal to ‘N’ else return true at the end of the loops.

Below is the implementation of the above approach:

C++

// C++ implementation for the
// above approach

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

// Function to check if number
// is an Idoneal numbers
bool isIdoneal(int n)
{
    // iterate for all
    // triples pairs (a, b, c)
    for (int a = 1; a <= n; a++) {
        for (int b = a + 1; b <= n; b++) {
            for (int c = b + 1; c <= n; c++) {

                // if the condition
                // is satisfied
                if (a * b + b * c + c * a == n)
                    return false;
            }
        }
    }
    return true;
}

// Driver Code
int main()
{
    int N = 10;

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

Java

// Java implementation for the
// above approach
import java.lang.*;

class GFG{

// Function to check if number
// is an Idoneal numbers
static boolean isIdoneal(int n)
{

    // Iterate for all
    // triples pairs (a, b, c)
    for(int a = 1; a <= n; a++)
    {
       for(int b = a + 1; b <= n; b++)
       {
          for(int c = b + 1; c <= n; c++)
          {

             // If the condition
             // is satisfied
             if (a * b + b * c + c * a == n)
                 return false;
          }
       }
    }
    return true;
}

// Driver Code
public static void main(String[] args)
{
    int N = 10;

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

// This code is contributed by rock_cool

Python3

# Python3 implementation for the
# above approach

# Function to check if number
# is an Idoneal numbers
def isIdoneal(n):

    # Iterate for all
    # triples pairs (a, b, c)
    for a in range(1, n + 1):
        for b in range(a + 1, n + 1):
            for c in range(b + 1, n + 1):

                # If the condition
                # is satisfied
                if (a * b + b * c + c * a == n):
                    return False

    return True

# Driver Code
N = 10

# Function call
if (isIdoneal(N)):
    print("Yes")
else:
    print("No")

# This code is contributed by Vishal Maurya.

C#

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

// Function to check if number
// is an Idoneal numbers
static bool isIdoneal(int n)
{

    // Iterate for all
    // triples pairs (a, b, c)
    for(int a = 1; a <= n; a++)
    {
        for(int b = a + 1; b <= n; b++)
        {
            for(int c = b + 1; c <= n; c++)
            {

                // If the condition
                // is satisfied
                if (a * b + b * c + c * a == n)
                    return false;
            }
        }
    }
    return true;
}

// Driver Code
public static void Main()
{
    int N = 10;

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

// This code is contributed by Code_Mech

Javascript

<script>

// Javascript implementation for the
// above approach

// Function to check if number
// is an Idoneal numbers
function isIdoneal(n)
{

    // iterate for all
    // triples pairs (a, b, c)
    for (var a = 1; a <= n; a++) {
        for (var b = a + 1; b <= n; b++) {
            for (var c = b + 1; c <= n; c++) {

                // if the condition
                // is satisfied
                if (a * b + b * c + c * a == n)
                    return false;
            }
        }
    }
    return true;
}

// Driver Code
var N = 10;

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

// This code is contributed by rutvik_56.
</script>

Output:

Yes

Time Complexity: O(n*n*n) 
Auxiliary Space: O(1) 

Reference: https://en.wikipedia.org/wiki/Idoneal_number