3-digit Osiris number

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Given a 3-digit number N, the task is to find if N is an Osiris number or not. Osiris numbers are the numbers that are equal to the sum of permutations of sub-samples of their own digits. For example, 132 is an Osiris number as it is equal to 12 + 21 + 13 + 31 + 23 + 32.

Examples:

Input: N = 132
Output: Yes
12 + 21 + 13 + 31 + 23 + 32 = 132

Input: N = 154
Output: No

Approach:

If n = 132,
132 = 12 + 21 + 13 + 31 + 23 + 32
132 = 2 * 11 + 2 * 22 + 2 * 33
132 = 22 + 44 + 66
132 = (2 + 4 + 6) * 11
132 = 2 * (1 + 2 + 3) * 11, each digit of 132 occurs twice in the ones and tens position of the sums.
The same rule applies for every 3-digit Osiris number and can be reciprocated to check whether a number is an Osiris number or not.
For a 3-digit number N to be considered as an Osiris number, N must be equal to 2 * (sum of digits) * 11

Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function that returns true if
// n is an Osiris number
bool isOsiris(int n)
{
    // 3rd digit
    int a = n % 10;
 
    // 2nd digit
    int b = (n / 10) % 10;
 
    // 1st digit
    int c = n / 100;
 
    int digit_sum = a + b + c;
 
    // Check the required condition
    if (n == (2 * (digit_sum)*11)) {
        return true;
    }
 
    return false;
}
 
// Driver code
int main()
{
    int n = 132;
    if (isOsiris(n))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
     
// Function that returns true if
// n is an Osiris number
static boolean isOsiris(int n)
{
    // 3rd digit
    int a = n % 10;
 
    // 2nd digit
    int b = (n / 10) % 10;
 
    // 1st digit
    int c = n / 100;
 
    int digit_sum = a + b + c;
 
    // Check the required condition
    if (n == (2 * (digit_sum)*11)) 
    {
        return true;
    }
 
    return false;
}
 
// Driver code
public static void main(String args[])
{
    int n = 132;
    if (isOsiris(n))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by Akanksha Rai

Python3

# Python implementation of the approach
 
# Function that returns true if 
# n is an Osiris number
def isOsiris(n):
     
    # 3rd digit 
    a = n % 10
     
    # 2nd digit
    b = (n//10)% 10
     
    # 1st digit
    c = n//100
 
    digit_sum = a + b + c
 
    # Check the required condition
    if(n == (2 * (digit_sum) * 11)):
        return True
     
    return False
 
# Driver code
if __name__ == '__main__':
    n = 132
    if isOsiris(n):
        print("Yes")
    else :
        print("No")

C#

// C# implementation of the approach
using System;
 
class GFG
{
// Function that returns true if
// n is an Osiris number
static bool isOsiris(int n)
{
    // 3rd digit
    int a = n % 10;
 
    // 2nd digit
    int b = (n / 10) % 10;
 
    // 1st digit
    int c = n / 100;
 
    int digit_sum = a + b + c;
 
    // Check the required condition
    if (n == (2 * (digit_sum)*11)) 
    {
        return true;
    }
 
    return false;
}
 
// Driver code
static void Main()
{
    int n = 132;
    if (isOsiris(n))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by mits

PHP

<?php
// PHP implementation of the approach 
 
// Function that returns true if 
// n is an Osiris number 
function isOsiris($n) 
{ 
    // 3rd digit 
    $a = $n % 10; 
 
    // 2nd digit 
    $b = floor($n / 10) % 10; 
 
    // 1st digit 
    $c = floor($n / 100); 
 
    $digit_sum = $a + $b + $c; 
 
    // Check the required condition 
    if ($n == (2 * ($digit_sum) * 11))
    { 
        return true; 
    } 
 
    return false; 
} 
 
// Driver code 
$n = 132; 
if (isOsiris($n)) 
    echo "Yes"; 
else
    echo "No"; 
     
// This code is contributed by Ryuga
?>

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function that returns true if
// n is an Osiris number
function isOsiris(n)
{
     
    // 3rd digit
    let a = n % 10;
 
    // 2nd digit
    let b = parseInt((n / 10) % 10);
 
    // 1st digit
    let c = parseInt(n / 100);
 
    let digit_sum = a + b + c;
 
    // Check the required condition
    if (n == (2 * (digit_sum) * 11))
    {
        return true;
    }
    return false;
}
 
// Driver code
let n = 132;
 
if (isOsiris(n))
    document.write("Yes");
else
    document.write("No");
     
// This code is contributed by souravmahato348
 
</script>

Output:

Yes

Time Complexity: O(1)
Space Complexity: O(1)

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