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Find all array elements whose digit sum are equal to perfect number

The problem at hand involves finding array elements whose digit sum equals a perfect number. A perfect number is a positive integer that is equal to the sum of its proper divisors, excluding itself. For example, the number 28 is a perfect number because its divisors (excluding itself) are 1, 2, 4, 7, and 14, and their sum is indeed 28.

Given an array of integers, we need to identify and display those elements whose individual digits sum up to a perfect number. This requires breaking down each integer into its digits, summing them up, and then checking if the resulting sum is a perfect number.

Problem Statement

The problem can be broken down into a few key steps:

  1. Iterate through the array of integers.
  2. For each integer, calculate the sum of its digits.
  3. Check if the calculated digit sum is a perfect number.
  4. If it's a perfect number, display the integer.

Explanation with Example

Let's consider an example to illustrate the problem. We have the following array: arr = [5, 231, 4, 60, -34, 4888].

For each integer in the array, we calculate the digit sum:

  • 5 → Digit sum: 5
  • 231 → Digit sum: 2 + 3 + 1 = 6
  • 4 → Digit sum: 4
  • 60 → Digit sum: 6 + 0 = 6
  • -34 → Ignoring negative integers
  • 4888 → Digit sum: 4 + 8 + 8 + 8 = 28

We observe that the digit sums of 231 and 60 are perfect numbers (6), and the digit sum of 4888 is also a perfect number (28).

Idea to Solve the Problem

To solve this problem, we need to follow these steps:

  1. Define a function to calculate the digit sum of an integer.
  2. Define a function to check if a given number is a perfect number.
  3. Iterate through the array of integers and check if their digit sums are perfect numbers.

Pseudocode

Function digitSum(num):
    n = num
    result = 0
    while n is not 0:
        result += n % 10
        n = n / 10
    return result

Function isPerfectNo(n):
    if n < 0:
        return false
    sum = 0
    for i from n/2 down to 1:
        if n % i is 0:
            sum += i
    return sum is n

Function findPerfectNo(arr, n):
    result = false
    for i from 0 to n - 1:
        if arr[i] > 0 and isPerfectNo(digitSum(arr[i])):
            print arr[i]
            result = true
    if result is false:
        print "None"

Main:
    arr = [5, 231, 4, 60, -34, 4888]
    n = length of arr
    findPerfectNo(arr, n)

Algorithm Explanation

  1. The digitSum function takes an integer as input and calculates the sum of its digits.
  2. The isPerfectNo function checks if a given number is a perfect number by iterating through its divisors.
  3. The findPerfectNo function iterates through the array, checks if the digit sum is a perfect number, and prints the result accordingly.
  4. In the Main section, we initialize the array arr, find its length n, and call the findPerfectNo function.

Code Solution

/*
    Java program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
public class PerfectNumber
{
    public int digitSum(int num)
    {
        int n = num;

        int result = 0;

        // Calculate digit sum
        while (n != 0)
        {
            // Add last digit
            result += (n % 10);

            // Remove last digit
            n = n / 10;
        }
        return result;
    }
    // Check given number is perfect or not
    public boolean isPerfectNo(int n)
    {
        if (n < 0)
        {
            return false;
        }
        int sum = 0;

        // Calculate sum of divisors of given n
        for (int i = (n / 2); i >= 1; i--)
        {
            if (n % i == 0)
            {
                sum += i;
            }
        }
        if (sum == n)
        {
            // When number is perfect number
            return true;
        }
        else
        {
            return false;
        }
    }
    public void findPerfectNo(int[] arr, int n)
    {
        boolean result = false;

        for (int i = 0; i < n; ++i)
        {
            if (arr[i] > 0 && isPerfectNo(digitSum(arr[i])))
            {
                // When array element is perfect number
                System.out.print("  " + arr[i]);

                // Active result indicator
                result = true;
            }
        }
        if (result == false)
        {
            System.out.print("\n None \n");
        }
    }
    public static void main(String[] args)
    {
        PerfectNumber task = new PerfectNumber();
        int[] arr = {
            5 , 231 , 4 , 60 , -34 , 4888
        };
        int n = arr.length;
        /*
            arr = [5, 231, 4, 60, -34, 4888]
            --------------------------------
            2 + 3 + 1     = 6
            6 + 0         = 6
            4 + 8 + 8 + 8 = 28
            
            Result : [231,60,4888]
        */
        task.findPerfectNo(arr, n);
    }
}

Output

  231  60  4888
/*
    C program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
#include <stdio.h>


    int digitSum(int num)
    {
        int n = num;

        int result = 0;

        // Calculate digit sum
        while (n != 0)
        {
            // Add last digit
            result += (n % 10);

            // Remove last digit
            n = n / 10;
        }
        return result;
    }
    // Check given number is perfect or not
    int isPerfectNo(int n)
    {
        if (n < 0)
        {
            return 0;
        }
        int sum = 0;

        // Calculate sum of divisors of given n
        for (int i = (n / 2); i >= 1; i--)
        {
            if (n % i == 0)
            {
                sum += i;
            }
        }
        if (sum == n)
        {
            // When number is perfect number
            return 1;
        }
        else
        {
            return 0;
        }
    }
    void findPerfectNo(int arr[], int n)
    {
        int result = 0;

        for (int i = 0; i < n; ++i)
        {
            if (arr[i] > 0 && isPerfectNo(digitSum(arr[i])))
            {
                // When array element is perfect number
                printf("  %d", arr[i]);

                // Active result indicator
                result = 1;
            }
        }
        if (result == 0)
        {
            printf("\n None \n");
        }
    }
int main(int argc, char const *argv[])
{
        int arr[] = 
        {
            5 , 231 , 4 , 60 , -34 , 4888
        };
        int n = sizeof(arr) / sizeof(arr[0]);
        /*
            arr = [5, 231, 4, 60, -34, 4888]
            --------------------------------
            2 + 3 + 1     = 6
            6 + 0         = 6
            4 + 8 + 8 + 8 = 28
            
            Result : [231,60,4888]
        */
        findPerfectNo(arr, n);    
        return 0;
}

Output

  231  60  4888
// Include header file
#include <iostream>

using namespace std;
/*
    C++ program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
class PerfectNumber
{
	public: int digitSum(int num)
	{
		int n = num;
		int result = 0;
		// Calculate digit sum
		while (n != 0)
		{
			// Add last digit
			result += (n % 10);
			// Remove last digit
			n = n / 10;
		}
		return result;
	}
	// Check given number is perfect or not
	bool isPerfectNo(int n)
	{
		if (n < 0)
		{
			return false;
		}
		int sum = 0;
		// Calculate sum of divisors of given n
		for (int i = (n / 2); i >= 1; i--)
		{
			if (n % i == 0)
			{
				sum += i;
			}
		}
		if (sum == n)
		{
			// When number is perfect number
			return true;
		}
		else
		{
			return false;
		}
	}
	void findPerfectNo(int arr[], int n)
	{
		bool result = false;
		for (int i = 0; i < n; ++i)
		{
			if (arr[i] > 0 && 
                this->isPerfectNo(this->digitSum(arr[i])))
			{
				// When array element is perfect number
				cout << "  " << arr[i];
				// Active result indicator
				result = true;
			}
		}
		if (result == false)
		{
			cout << "\n None \n";
		}
	}
};
int main()
{
	PerfectNumber *task = new PerfectNumber();
	int arr[] = {
		5 , 231 , 4 , 60 , -34 , 4888
	};
	int n = sizeof(arr) / sizeof(arr[0]);
	/*
	    arr = [5, 231, 4, 60, -34, 4888]
	    --------------------------------
	    2 + 3 + 1     = 6
	    6 + 0         = 6
	    4 + 8 + 8 + 8 = 28

	    Result : [231,60,4888]
	        
	*/
	task->findPerfectNo(arr, n);
	return 0;
}

Output

  231  60  4888
// Include namespace system
using System;
/*
    Csharp program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
public class PerfectNumber
{
	public int digitSum(int num)
	{
		int n = num;
		int result = 0;
		// Calculate digit sum
		while (n != 0)
		{
			// Add last digit
			result += (n % 10);
			// Remove last digit
			n = n / 10;
		}
		return result;
	}
	// Check given number is perfect or not
	public Boolean isPerfectNo(int n)
	{
		if (n < 0)
		{
			return false;
		}
		int sum = 0;
		// Calculate sum of divisors of given n
		for (int i = (n / 2); i >= 1; i--)
		{
			if (n % i == 0)
			{
				sum += i;
			}
		}
		if (sum == n)
		{
			// When number is perfect number
			return true;
		}
		else
		{
			return false;
		}
	}
	public void findPerfectNo(int[] arr, int n)
	{
		Boolean result = false;
		for (int i = 0; i < n; ++i)
		{
			if (arr[i] > 0 && 
                this.isPerfectNo(this.digitSum(arr[i])))
			{
				// When array element is perfect number
				Console.Write("  " + arr[i]);
				// Active result indicator
				result = true;
			}
		}
		if (result == false)
		{
			Console.Write("\n None \n");
		}
	}
	public static void Main(String[] args)
	{
		PerfectNumber task = new PerfectNumber();
		int[] arr = {
			5 , 231 , 4 , 60 , -34 , 4888
		};
		int n = arr.Length;
		/*
		    arr = [5, 231, 4, 60, -34, 4888]
		    --------------------------------
		    2 + 3 + 1     = 6
		    6 + 0         = 6
		    4 + 8 + 8 + 8 = 28

		    Result : [231,60,4888]
		        
		*/
		task.findPerfectNo(arr, n);
	}
}

Output

  231  60  4888
package main
import "fmt"
/*
    Go program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/

func digitSum(num int) int {
	var n int = num
	var result int = 0
	// Calculate digit sum
	for (n != 0) {
		// Add last digit
		result += (n % 10)
		// Remove last digit
		n = n / 10
	}
	return result
}
// Check given number is perfect or not
func isPerfectNo(n int) bool {
	if n < 0 {
		return false
	}
	var sum int = 0
	// Calculate sum of divisors of given n
	for i :=(n / 2) ; i >= 1 ; i-- {
		if n % i == 0 {
			sum += i
		}
	}
	if sum == n {
		// When number is perfect number
		return true
	} else {
		return false
	}
}
func findPerfectNo(arr[] int, n int) {
	var result bool = false
	for i := 0 ; i < n ; i++ {
		if arr[i] > 0 && 
			isPerfectNo(digitSum(arr[i])) {
			// When array element is perfect number
			fmt.Print("  ", arr[i])
			// Active result indicator
			result = true
		}
	}
	if result == false {
		fmt.Print("\n None \n")
	}
}
func main() {
	
	var arr = [] int { 5 , 231 , 4 , 60 , - 34 , 4888}
	var n int = len(arr)
	/*
	    arr = [5, 231, 4, 60, -34, 4888]
	    --------------------------------
	    2 + 3 + 1     = 6
	    6 + 0         = 6
	    4 + 8 + 8 + 8 = 28

	    Result : [231,60,4888]
	        
	*/
	findPerfectNo(arr, n)
}

Output

  231  60  4888
<?php
/*
    Php program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
class PerfectNumber
{
	public	function digitSum($num)
	{
		$n = $num;
		$result = 0;
		// Calculate digit sum
		while ($n != 0)
		{
			// Add last digit
			$result += ($n % 10);
			// Remove last digit
			$n = (int)($n / 10);
		}
		return $result;
	}
	// Check given number is perfect or not
	public	function isPerfectNo($n)
	{
		if ($n < 0)
		{
			return false;
		}
		$sum = 0;
		// Calculate sum of divisors of given n
		for ($i = ((int)($n / 2)); $i >= 1; $i--)
		{
			if ($n % $i == 0)
			{
				$sum += $i;
			}
		}
		if ($sum == $n)
		{
			// When number is perfect number
			return true;
		}
		else
		{
			return false;
		}
	}
	public	function findPerfectNo($arr, $n)
	{
		$result = false;
		for ($i = 0; $i < $n; ++$i)
		{
			if ($arr[$i] > 0 && 
                $this->isPerfectNo($this->digitSum($arr[$i])))
			{
				// When array element is perfect number
				echo("  ".$arr[$i]);
				// Active result indicator
				$result = true;
			}
		}
		if ($result == false)
		{
			echo("\n None \n");
		}
	}
}

function main()
{
	$task = new PerfectNumber();
	$arr = array(5, 231, 4, 60, -34, 4888);
	$n = count($arr);
	/*
	    arr = [5, 231, 4, 60, -34, 4888]
	    --------------------------------
	    2 + 3 + 1     = 6
	    6 + 0         = 6
	    4 + 8 + 8 + 8 = 28

	    Result : [231,60,4888]
	        
	*/
	$task->findPerfectNo($arr, $n);
}
main();

Output

  231  60  4888
/*
    Node JS program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
class PerfectNumber
{
	digitSum(num)
	{
		var n = num;
		var result = 0;
		// Calculate digit sum
		while (n != 0)
		{
			// Add last digit
			result += (n % 10);
			// Remove last digit
			n = parseInt(n / 10);
		}
		return result;
	}
	// Check given number is perfect or not
	isPerfectNo(n)
	{
		if (n < 0)
		{
			return false;
		}
		var sum = 0;
		// Calculate sum of divisors of given n
		for (var i = (parseInt(n / 2)); i >= 1; i--)
		{
			if (n % i == 0)
			{
				sum += i;
			}
		}
		if (sum == n)
		{
			// When number is perfect number
			return true;
		}
		else
		{
			return false;
		}
	}
	findPerfectNo(arr, n)
	{
		var result = false;
		for (var i = 0; i < n; ++i)
		{
			if (arr[i] > 0 && 
                this.isPerfectNo(this.digitSum(arr[i])))
			{
				// When array element is perfect number
				process.stdout.write("  " + arr[i]);
				// Active result indicator
				result = true;
			}
		}
		if (result == false)
		{
			process.stdout.write("\n None \n");
		}
	}
}

function main()
{
	var task = new PerfectNumber();
	var arr = [5, 231, 4, 60, -34, 4888];
	var n = arr.length;
	/*
	    arr = [5, 231, 4, 60, -34, 4888]
	    --------------------------------
	    2 + 3 + 1     = 6
	    6 + 0         = 6
	    4 + 8 + 8 + 8 = 28

	    Result : [231,60,4888]
	        
	*/
	task.findPerfectNo(arr, n);
}
main();

Output

  231  60  4888
#    Python 3 program for
#    Find all array elements whose digit sum 
#    are equal to perfect number
class PerfectNumber :
	def digitSum(self, num) :
		n = num
		result = 0
		#  Calculate digit sum
		while (n != 0) :
			#  Add last digit
			result += (n % 10)
			#  Remove last digit
			n = int(n / 10)
		
		return result
	
	#  Check given number is perfect or not
	def isPerfectNo(self, n) :
		if (n < 0) :
			return False
		
		sum = 0
		i = (int(n / 2))
		#  Calculate sum of divisors of given n
		while (i >= 1) :
			if (n % i == 0) :
				sum += i
			
			i -= 1
		
		if (sum == n) :
			#  When number is perfect number
			return True
		else :
			return False
		
	
	def findPerfectNo(self, arr, n) :
		result = False
		i = 0
		while (i < n) :
			if (arr[i] > 0 and self.isPerfectNo(self.digitSum(arr[i]))) :
				#  When list element is perfect number
				print("  ", arr[i], end = "")
				#  Active result indicator
				result = True
			
			i += 1
		
		if (result == False) :
			print("\n None ")
		
	

def main() :
	task = PerfectNumber()
	arr = [5, 231, 4, 60, -34, 4888]
	n = len(arr)
	#    arr = [5, 231, 4, 60, -34, 4888]
	#    --------------------------------
	#    2 + 3 + 1     = 6
	#    6 + 0         = 6
	#    4 + 8 + 8 + 8 = 28
	#    Result : [231,60,4888]
	task.findPerfectNo(arr, n)

if __name__ == "__main__": main()

Output

   231   60   4888
#    Ruby program for
#    Find all array elements whose digit sum 
#    are equal to perfect number
class PerfectNumber 
	def digitSum(num) 
		n = num
		result = 0
		#  Calculate digit sum
		while (n != 0) 
			#  Add last digit
			result += (n % 10)
			#  Remove last digit
			n = n / 10
		end

		return result
	end

	#  Check given number is perfect or not
	def isPerfectNo(n) 
		if (n < 0) 
			return false
		end

		sum = 0
		i = (n / 2)
		#  Calculate sum of divisors of given n
		while (i >= 1) 
			if (n % i == 0) 
				sum += i
			end

			i -= 1
		end

		if (sum == n) 
			#  When number is perfect number
			return true
		else
 
			return false
		end

	end

	def findPerfectNo(arr, n) 
		result = false
		i = 0
		while (i < n) 
			if (arr[i] > 0 && self.isPerfectNo(self.digitSum(arr[i]))) 
				#  When array element is perfect number
				print("  ", arr[i])
				#  Active result indicator
				result = true
			end

			i += 1
		end

		if (result == false) 
			print("\n None \n")
		end

	end

end

def main() 
	task = PerfectNumber.new()
	arr = [5, 231, 4, 60, -34, 4888]
	n = arr.length
	#    arr = [5, 231, 4, 60, -34, 4888]
	#    --------------------------------
	#    2 + 3 + 1     = 6
	#    6 + 0         = 6
	#    4 + 8 + 8 + 8 = 28
	#    Result : [231,60,4888]
	task.findPerfectNo(arr, n)
end

main()

Output

  231  60  4888
/*
    Scala program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
class PerfectNumber()
{
	def digitSum(num: Int): Int = {
		var n: Int = num;
		var result: Int = 0;
		// Calculate digit sum
		while (n != 0)
		{
			// Add last digit
			result += (n % 10);
			// Remove last digit
			n = n / 10;
		}
		return result;
	}
	// Check given number is perfect or not
	def isPerfectNo(n: Int): Boolean = {
		if (n < 0)
		{
			return false;
		}
		var sum: Int = 0;
		var i: Int = (n / 2);
		// Calculate sum of divisors of given n
		while (i >= 1)
		{
			if (n % i == 0)
			{
				sum += i;
			}
			i -= 1;
		}
		if (sum == n)
		{
			// When number is perfect number
			return true;
		}
		else
		{
			return false;
		}
	}
	def findPerfectNo(arr: Array[Int], n: Int): Unit = {
		var result: Boolean = false;
		var i: Int = 0;
		while (i < n)
		{
			if (arr(i) > 0 && isPerfectNo(digitSum(arr(i))))
			{
				// When array element is perfect number
				print("  " + arr(i));
				// Active result indicator
				result = true;
			}
			i += 1;
		}
		if (result == false)
		{
			print("\n None \n");
		}
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: PerfectNumber = new PerfectNumber();
		var arr: Array[Int] = Array(5, 231, 4, 60, -34, 4888);
		var n: Int = arr.length;
		/*
		    arr = [5, 231, 4, 60, -34, 4888]
		    --------------------------------
		    2 + 3 + 1     = 6
		    6 + 0         = 6
		    4 + 8 + 8 + 8 = 28

		    Result : [231,60,4888]
		        
		*/
		task.findPerfectNo(arr, n);
	}
}

Output

  231  60  4888
import Foundation;
/*
    Swift 4 program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
class PerfectNumber
{
	func digitSum(_ num: Int) -> Int
	{
		var n: Int = num;
		var result: Int = 0;
		// Calculate digit sum
		while (n  != 0)
		{
			// Add last digit
			result += (n % 10);
			// Remove last digit
			n = n / 10;
		}
		return result;
	}
	// Check given number is perfect or not
	func isPerfectNo(_ n: Int) -> Bool
	{
		if (n < 0)
		{
			return false;
		}
		var sum: Int = 0;
		var i: Int = (n / 2);
		// Calculate sum of divisors of given n
		while (i >= 1)
		{
			if (n % i == 0)
			{
				sum += i;
			}
			i -= 1;
		}
		if (sum == n)
		{
			// When number is perfect number
			return true;
		}
		else
		{
			return false;
		}
	}
	func findPerfectNo(_ arr: [Int], _ n: Int)
	{
		var result: Bool = false;
		var i: Int = 0;
		while (i < n)
		{
			if (arr[i] > 0 && self.isPerfectNo(self.digitSum(arr[i])))
			{
				// When array element is perfect number
				print("  ", arr[i], terminator: "");
				// Active result indicator
				result = true;
			}
			i += 1;
		}
		if (result == false)
		{
			print("\n None ");
		}
	}
}
func main()
{
	let task: PerfectNumber = PerfectNumber();
	let arr: [Int] = [5, 231, 4, 60, -34, 4888];
	let n: Int = arr.count;
	/*
	    arr = [5, 231, 4, 60, -34, 4888]
	    --------------------------------
	    2 + 3 + 1     = 6
	    6 + 0         = 6
	    4 + 8 + 8 + 8 = 28

	    Result : [231,60,4888]
	        
	*/
	task.findPerfectNo(arr, n);
}
main();

Output

   231   60   4888
/*
    Kotlin program for
    Find all array elements whose digit sum 
    are equal to perfect number
*/
class PerfectNumber
{
	fun digitSum(num: Int): Int
	{
		var n: Int = num;
		var result: Int = 0;
		// Calculate digit sum
		while (n != 0)
		{
			// Add last digit
			result += (n % 10);
			// Remove last digit
			n = n / 10;
		}
		return result;
	}
	// Check given number is perfect or not
	fun isPerfectNo(n: Int): Boolean
	{
		if (n < 0)
		{
			return false;
		}
		var sum: Int = 0;
		var i: Int = (n / 2);
		// Calculate sum of divisors of given n
		while (i >= 1)
		{
			if (n % i == 0)
			{
				sum += i;
			}
			i -= 1;
		}
		if (sum == n)
		{
			// When number is perfect number
			return true;
		}
		else
		{
			return false;
		}
	}
	fun findPerfectNo(arr: Array < Int > , n: Int): Unit
	{
		var result: Boolean = false;
		var i: Int = 0;
		while (i < n)
		{
			if (arr[i] > 0 && this.isPerfectNo(this.digitSum(arr[i])))
			{
				// When array element is perfect number
				print("  " + arr[i]);
				// Active result indicator
				result = true;
			}
			i += 1;
		}
		if (result == false)
		{
			print("\n None \n");
		}
	}
}
fun main(args: Array < String > ): Unit
{
	val task: PerfectNumber = PerfectNumber();
	val arr: Array < Int > = arrayOf(5, 231, 4, 60, -34, 4888);
	val n: Int = arr.count();
	/*
	    arr = [5, 231, 4, 60, -34, 4888]
	    --------------------------------
	    2 + 3 + 1     = 6
	    6 + 0         = 6
	    4 + 8 + 8 + 8 = 28

	    Result : [231,60,4888]
	        
	*/
	task.findPerfectNo(arr, n);
}

Output

  231  60  4888

Time Complexity Analysis

  • The digitSum function has a time complexity of O(log n), where n is the input number, as it involves dividing the number by 10 in each iteration of the loop.
  • The isPerfectNo function has a time complexity of O(n), where n is the input number, as it involves iterating through the divisors up to n/2.
  • The findPerfectNo function has a time complexity of O(m * (log n + n/2)), where m is the number of elements in the array and n is the maximum value among the array elements. This is because it calls both digitSum and isPerfectNo functions for each array element.
  • Thus, the overall time complexity of the given code is dominated by the findPerfectNo function, which is O(m * (log n + n/2)).
Last updated on by Kalkicode




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