Posted on by Kalkicode
Code Mathematics

Find two numbers whose sum is N and does not contain any digit as K

The statement "Find two numbers whose sum is N and does not contain any digit as K" means that you need to find two whole numbers that add up to a given number N, and neither of the numbers can contain the digit K.

For example, if N is 15 and K is 5, you would need to find two numbers that add up to 15 and do not contain the digit 5. One possible solution would be 8 and 7, since 8 does not contain the digit 5 and 7 does not contain the digit 5 either.

Note that when finding these numbers, it is important to consider all possible numbers that can be formed using the digits that are not equal to K. Additionally, it is important to ensure that both numbers are whole numbers and that they add up to N.

Program Soltution

// C Program
// Find two numbers whose sum is N and does not contain any digit as K
#include <stdio.h>

int checkDigitK(int number, int k)
{
	int temp = number;
	while (temp != 0)
	{
		if (temp % 10 == k)
		{
			// When number contains k digit
			return 1;
		}
		temp = temp / 10;
	}
	return 0;
}
void constructWithoutK(int n, int k)
{
	if (n <= 0 || (k < 0 || k > 9))
	{
		return;
	}
	int temp = n;
	int extra = 0;
	int count = 1;
	while (temp != 0)
	{
		if (temp % 10 == k)
		{
			extra += count;
		}
		temp = temp / 10;
		count = count *10;
	}
	printf("\n Given sum : %d ", n);
	printf("\n Given k   : %d ", k);
	if (extra == 0 && !checkDigitK(n, k))
	{
		printf("\n %d (%d + %d) \n", n, 0, n);
		return;
	}
	temp = n - extra;
	if (extra > temp)
	{
		// Swap value
		count = extra;
		extra = temp;
		temp = count;
	}
	while (extra < temp)
	{
		if (!checkDigitK(temp, k) && 
            !checkDigitK(extra, k))
		{
			printf("\n %d (%d + %d) \n", n, temp, extra);
			return;
		}
		extra++;
		temp--;
	}
	printf("\n number :  Without digit %d value %d not possible \n", n, k);
}
int main()
{
	// Test 
	constructWithoutK(111, 1);
	constructWithoutK(1, 1);
	constructWithoutK(123, 2);
	constructWithoutK(7331, 7);
	constructWithoutK(333, 3);
	return 0;
}

Output

 Given sum : 111
 Given k   : 1
 111 (89 + 22)

 Given sum : 1
 Given k   : 1
 number :  Without digit 1 value 1 not possible

 Given sum : 123
 Given k   : 2
 123 (113 + 10)

 Given sum : 7331
 Given k   : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k   : 3
 333 (222 + 111)
// Java program for
// Find two numbers whose sum is N and does not contain any digit as K
public class Construction
{
    public boolean checkDigitK(int number, int k)
    {
        int temp = number;
        while (temp != 0)
        {
            if (temp % 10 == k)
            {
                // When number contains k digit
                return true;
            }
            temp = temp / 10;
        }
        return false;
    }
    public void constructWithoutK(int n, int k)
    {
        if (n <= 0 || (k < 0 || k > 9))
        {
            return;
        }
        int temp = n;
        int extra = 0;
        int count = 1;
        while (temp != 0)
        {
            if (temp % 10 == k)
            {
                extra += count;
            }
            temp = temp / 10;
            count = count * 10;
        }
        System.out.print("\n Given sum : " + n);
        System.out.print("\n Given k : " + k);
        if (extra == 0 && !checkDigitK(n, k))
        {
            System.out.print("\n " + n + " (" + 0 + " + " + n + ") \n");
            return;
        }
        temp = n - extra;
        if (extra > temp)
        {
            // Swap value
            count = extra;
            extra = temp;
            temp = count;
        }
        while (extra < temp)
        {
            if (!checkDigitK(temp, k) && !checkDigitK(extra, k))
            {
                System.out.print("\n " + n + " (" + temp + " + " + extra + ") \n");
                return;
            }
            extra++;
            temp--;
        }
        System.out.print("\n number : Without digit " + n + " value " + k + " not possible \n");
    }
    public static void main(String[] args)
    {
        Construction task = new Construction();
        // Test  A
        // Number : 111
        // K      :  1
        task.constructWithoutK(111, 1);
        // Test  B
        // Number :  1
        // K      :  1
        task.constructWithoutK(1, 1);
        // Test  C
        // Number :  123
        // K      :  2
        task.constructWithoutK(123, 2);
        // Test  D
        // Number :  7331
        // K      :  7
        task.constructWithoutK(7331, 7);
        // Test  E
        // Number :  333
        // K      :  3
        task.constructWithoutK(333, 3);
    }
}

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)
// Include header file
#include <iostream>
using namespace std;
// C++ program for
// Find two numbers whose sum is N and 
// does not contain any digit as K
class Construction
{
	public: bool checkDigitK(int number, int k)
	{
		int temp = number;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				// When number contains k digit
				return true;
			}
			temp = temp / 10;
		}
		return false;
	}
	void constructWithoutK(int n, int k)
	{
		if (n <= 0 || (k < 0 || k > 9))
		{
			return;
		}
		int temp = n;
		int extra = 0;
		int count = 1;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				extra += count;
			}
			temp = temp / 10;
			count = count *10;
		}
		cout << "\n Given sum : " << n;
		cout << "\n Given k : " << k;
		if (extra == 0 && !this->checkDigitK(n, k))
		{
			cout << "\n " << n << " (" << 0 << " + " << n << ") \n";
			return;
		}
		temp = n - extra;
		if (extra > temp)
		{
			// Swap value
			count = extra;
			extra = temp;
			temp = count;
		}
		while (extra < temp)
		{
			if (!this->checkDigitK(temp, k) && !this->checkDigitK(extra, k))
			{
				cout << "\n " << n << " (" 
              		 << temp << " + " << extra << ") \n";
				return;
			}
			extra++;
			temp--;
		}
		cout << "\n number : Without digit " 
             << n << " value " << k << " not possible \n";
	}
};
int main()
{
	Construction *task = new Construction();
	// Test  A
	// Number : 111
	// K      :  1
	task->constructWithoutK(111, 1);
	// Test  B
	// Number :  1
	// K      :  1
	task->constructWithoutK(1, 1);
	// Test  C
	// Number :  123
	// K      :  2
	task->constructWithoutK(123, 2);
	// Test  D
	// Number :  7331
	// K      :  7
	task->constructWithoutK(7331, 7);
	// Test  E
	// Number :  333
	// K      :  3
	task->constructWithoutK(333, 3);
	return 0;
}

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)
// Include namespace system
using System;
// Csharp program for
// Find two numbers whose sum is N 
// and does not contain any digit as K
public class Construction
{
	public Boolean checkDigitK(int number, int k)
	{
		int temp = number;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				// When number contains k digit
				return true;
			}
			temp = temp / 10;
		}
		return false;
	}
	public void constructWithoutK(int n, int k)
	{
		if (n <= 0 || (k < 0 || k > 9))
		{
			return;
		}
		int temp = n;
		int extra = 0;
		int count = 1;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				extra += count;
			}
			temp = temp / 10;
			count = count * 10;
		}
		Console.Write("\n Given sum : " + n);
		Console.Write("\n Given k : " + k);
		if (extra == 0 && !this.checkDigitK(n, k))
		{
			Console.Write("\n " + n + " (" + 0 + " + " + n + ") \n");
			return;
		}
		temp = n - extra;
		if (extra > temp)
		{
			// Swap value
			count = extra;
			extra = temp;
			temp = count;
		}
		while (extra < temp)
		{
			if (!this.checkDigitK(temp, k) && !this.checkDigitK(extra, k))
			{
				Console.Write("\n " + n + " (" + temp + " + " + extra + ") \n");
				return;
			}
			extra++;
			temp--;
		}
		Console.Write("\n number : Without digit " + n + " value " + k + " not possible \n");
	}
	public static void Main(String[] args)
	{
		Construction task = new Construction();
		// Test  A
		// Number : 111
		// K      :  1
		task.constructWithoutK(111, 1);
		// Test  B
		// Number :  1
		// K      :  1
		task.constructWithoutK(1, 1);
		// Test  C
		// Number :  123
		// K      :  2
		task.constructWithoutK(123, 2);
		// Test  D
		// Number :  7331
		// K      :  7
		task.constructWithoutK(7331, 7);
		// Test  E
		// Number :  333
		// K      :  3
		task.constructWithoutK(333, 3);
	}
}

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)
package main
import "fmt"
// Go program for
// Find two numbers whose sum is N 
// and does not contain any digit as K
type Construction struct {}
func getConstruction() * Construction {
	var me *Construction = &Construction {}
	return me
}
func(this Construction) checkDigitK(number, k int) bool {
	var temp int = number
	for (temp != 0) {
		if temp % 10 == k {
			// When number contains k digit
			return true
		}
		temp = temp / 10
	}
	return false
}
func(this Construction) constructWithoutK(n, k int) {
	if n <= 0 || (k < 0 || k > 9) {
		return
	}
	var temp int = n
	var extra int = 0
	var count int = 1
	for (temp != 0) {
		if temp % 10 == k {
			extra += count
		}
		temp = temp / 10
		count = count * 10
	}
	fmt.Print("\n Given sum : ", n)
	fmt.Print("\n Given k : ", k)
	if extra == 0 && !this.checkDigitK(n, k) {
		fmt.Print("\n ", n, " (", 0, " + ", n, ") \n")
		return
	}
	temp = n - extra
	if extra > temp {
		// Swap value
		count = extra
		extra = temp
		temp = count
	}
	for (extra < temp) {
		if !this.checkDigitK(temp, k) && 
		!this.checkDigitK(extra, k) {
			fmt.Print("\n ", n, " (", temp, " + ", extra, ") \n")
			return
		}
		extra++
		temp--
	}
	fmt.Print("\n number : Without digit ", n, 
		" value ", k, " not possible \n")
}
func main() {
	var task * Construction = getConstruction()
	// Test  A
	// Number : 111
	// K      :  1
	task.constructWithoutK(111, 1)
	// Test  B
	// Number :  1
	// K      :  1
	task.constructWithoutK(1, 1)
	// Test  C
	// Number :  123
	// K      :  2
	task.constructWithoutK(123, 2)
	// Test  D
	// Number :  7331
	// K      :  7
	task.constructWithoutK(7331, 7)
	// Test  E
	// Number :  333
	// K      :  3
	task.constructWithoutK(333, 3)
}

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)
<?php
// Php program for
// Find two numbers whose sum is N 
// and does not contain any digit as K
class Construction
{
	public	function checkDigitK($number, $k)
	{
		$temp = $number;
		while ($temp != 0)
		{
			if ($temp % 10 == $k)
			{
				// When number contains k digit
				return true;
			}
			$temp = (int)($temp / 10);
		}
		return false;
	}
	public	function constructWithoutK($n, $k)
	{
		if ($n <= 0 || ($k < 0 || $k > 9))
		{
			return;
		}
		$temp = $n;
		$extra = 0;
		$count = 1;
		while ($temp != 0)
		{
			if ($temp % 10 == $k)
			{
				$extra += $count;
			}
			$temp = (int)($temp / 10);
			$count = $count * 10;
		}
		echo("\n Given sum : ".$n);
		echo("\n Given k : ".$k);
		if ($extra == 0 && !$this->checkDigitK($n, $k))
		{
			echo("\n ".$n." (0 ".$n.") \n");
			return;
		}
		$temp = $n - $extra;
		if ($extra > $temp)
		{
			// Swap value
			$count = $extra;
			$extra = $temp;
			$temp = $count;
		}
		while ($extra < $temp)
		{
			if (!$this->checkDigitK($temp, $k) && 
                !$this->checkDigitK($extra, $k))
			{
				echo("\n ".$n." (".$temp." + ".$extra.") \n");
				return;
			}
			$extra++;
			$temp--;
		}
		echo("\n number : Without digit ".$n." value ".$k." not possible \n");
	}
}

function main()
{
	$task = new Construction();
	// Test  A
	// Number : 111
	// K      :  1
	$task->constructWithoutK(111, 1);
	// Test  B
	// Number :  1
	// K      :  1
	$task->constructWithoutK(1, 1);
	// Test  C
	// Number :  123
	// K      :  2
	$task->constructWithoutK(123, 2);
	// Test  D
	// Number :  7331
	// K      :  7
	$task->constructWithoutK(7331, 7);
	// Test  E
	// Number :  333
	// K      :  3
	$task->constructWithoutK(333, 3);
}
main();

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)
// Node JS program for
// Find two numbers whose sum is N 
// and does not contain any digit as K
class Construction
{
	checkDigitK(number, k)
	{
		var temp = number;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				// When number contains k digit
				return true;
			}
			temp = parseInt(temp / 10);
		}
		return false;
	}
	constructWithoutK(n, k)
	{
		if (n <= 0 || (k < 0 || k > 9))
		{
			return;
		}
		var temp = n;
		var extra = 0;
		var count = 1;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				extra += count;
			}
			temp = parseInt(temp / 10);
			count = count * 10;
		}
		process.stdout.write("\n Given sum : " + n);
		process.stdout.write("\n Given k : " + k);
		if (extra == 0 && !this.checkDigitK(n, k))
		{
			process.stdout.write("\n " + n + 
                                 " (" + 0 + " + " + 
                                 n + ") \n");
			return;
		}
		temp = n - extra;
		if (extra > temp)
		{
			// Swap value
			count = extra;
			extra = temp;
			temp = count;
		}
		while (extra < temp)
		{
			if (!this.checkDigitK(temp, k) && 
                !this.checkDigitK(extra, k))
			{
				process.stdout.write("\n " + n + 
                                     " (" + temp + " + " + 
                                     extra + ") \n");
				return;
			}
			extra++;
			temp--;
		}
		process.stdout.write("\n number : Without digit " + n + 
                             " value " + k + " not possible \n");
	}
}

function main()
{
	var task = new Construction();
	// Test  A
	// Number : 111
	// K      :  1
	task.constructWithoutK(111, 1);
	// Test  B
	// Number :  1
	// K      :  1
	task.constructWithoutK(1, 1);
	// Test  C
	// Number :  123
	// K      :  2
	task.constructWithoutK(123, 2);
	// Test  D
	// Number :  7331
	// K      :  7
	task.constructWithoutK(7331, 7);
	// Test  E
	// Number :  333
	// K      :  3
	task.constructWithoutK(333, 3);
}
main();

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)
#  Python 3 program for
#  Find two numbers whose sum is N 
#  and does not contain any digit as K
class Construction :
	def checkDigitK(self, number, k) :
		temp = number
		while (temp != 0) :
			if (temp % 10 == k) :
				#  When number contains k digit
				return True
			
			temp = int(temp / 10)
		
		return False
	
	def constructWithoutK(self, n, k) :
		if (n <= 0 or(k < 0 or k > 9)) :
			return
		
		temp = n
		extra = 0
		count = 1
		while (temp != 0) :
			if (temp % 10 == k) :
				extra += count
			
			temp = int(temp / 10)
			count = count * 10
		
		print("\n Given sum : ", n, end = "")
		print("\n Given k : ", k, end = "")
		if (extra == 0 and not self.checkDigitK(n, k)) :
			print("\n ", n ," (", 0 ,"+", n ,") ",sep="")
			return
		
		temp = n - extra
		if (extra > temp) :
			#  Swap value
			count = extra
			extra = temp
			temp = count
		
		while (extra < temp) :
			if (not self.checkDigitK(temp, k) and not 
            self.checkDigitK(extra, k)) :
				print("\n ", n ," (", temp ,"+", extra ,") ")
				return
			
			extra += 1
			temp -= 1
		
		print("\n number : Without digit ", n ," value ", k ," not possible ")
	

def main() :
	task = Construction()
	#  Test  A
	#  Number : 111
	#  K      :  1
	task.constructWithoutK(111, 1)
	#  Test  B
	#  Number :  1
	#  K      :  1
	task.constructWithoutK(1, 1)
	#  Test  C
	#  Number :  123
	#  K      :  2
	task.constructWithoutK(123, 2)
	#  Test  D
	#  Number :  7331
	#  K      :  7
	task.constructWithoutK(7331, 7)
	#  Test  E
	#  Number :  333
	#  K      :  3
	task.constructWithoutK(333, 3)

if __name__ == "__main__": main()

Output

 Given sum :  111
 Given k :  1
  111  ( 89 + 22 )

 Given sum :  1
 Given k :  1
 number : Without digit  1  value  1  not possible

 Given sum :  123
 Given k :  2
  123  ( 113 + 10 )

 Given sum :  7331
 Given k :  7
  7331  ( 6331 + 1000 )

 Given sum :  333
 Given k :  3
  333  ( 222 + 111 )
#  Ruby program for
#  Find two numbers whose sum is N 
#  and does not contain any digit as K
class Construction 
	def checkDigitK(number, k) 
		temp = number
		while (temp != 0) 
			if (temp % 10 == k) 
				#  When number contains k digit
				return true
			end

			temp = temp / 10
		end

		return false
	end

	def constructWithoutK(n, k) 
		if (n <= 0 || (k < 0 || k > 9)) 
			return
		end

		temp = n
		extra = 0
		count = 1
		while (temp != 0) 
			if (temp % 10 == k) 
				extra += count
			end

			temp = temp / 10
			count = count * 10
		end

		print("\n Given sum : ", n)
		print("\n Given k : ", k)
		if (extra == 0 && !self.checkDigitK(n, k)) 
			print("\n ", n ," (", 0 ," + ", n ,") \n")
			return
		end

		temp = n - extra
		if (extra > temp) 
			#  Swap value
			count = extra
			extra = temp
			temp = count
		end

		while (extra < temp) 
			if (!self.checkDigitK(temp, k) && 
                !self.checkDigitK(extra, k)) 
				print("\n ", n ," (", temp ," + ", extra ,") \n")
				return
			end

			extra += 1
			temp -= 1
		end

		print("\n number : Without digit ", n ,
              " value ", k ," not possible \n")
	end

end

def main() 
	task = Construction.new()
	#  Test  A
	#  Number : 111
	#  K      :  1
	task.constructWithoutK(111, 1)
	#  Test  B
	#  Number :  1
	#  K      :  1
	task.constructWithoutK(1, 1)
	#  Test  C
	#  Number :  123
	#  K      :  2
	task.constructWithoutK(123, 2)
	#  Test  D
	#  Number :  7331
	#  K      :  7
	task.constructWithoutK(7331, 7)
	#  Test  E
	#  Number :  333
	#  K      :  3
	task.constructWithoutK(333, 3)
end

main()

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22) 

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible 

 Given sum : 123
 Given k : 2
 123 (113 + 10) 

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000) 

 Given sum : 333
 Given k : 3
 333 (222 + 111) 

// Scala program for
// Find two numbers whose sum is N 
// and does not contain any digit as K
class Construction()
{
	def checkDigitK(number: Int, k: Int): Boolean = {
		var temp: Int = number;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				// When number contains k digit
				return true;
			}
			temp = temp / 10;
		}
		return false;
	}
	def constructWithoutK(n: Int, k: Int): Unit = {
		if (n <= 0 || (k < 0 || k > 9))
		{
			return;
		}
		var temp: Int = n;
		var extra: Int = 0;
		var count: Int = 1;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				extra += count;
			}
			temp = temp / 10;
			count = count * 10;
		}
		print("\n Given sum : " + n);
		print("\n Given k : " + k);
		if (extra == 0 && !checkDigitK(n, k))
		{
			print("\n " + n + " (" + 0 + " + " + n + ") \n");
			return;
		}
		temp = n - extra;
		if (extra > temp)
		{
			// Swap value
			count = extra;
			extra = temp;
			temp = count;
		}
		while (extra < temp)
		{
			if (!checkDigitK(temp, k) && !checkDigitK(extra, k))
			{
				print("\n " + n + " (" + temp + " + " + extra + ") \n");
				return;
			}
			extra += 1;
			temp -= 1;
		}
		print("\n number : Without digit " + n + 
              " value " + k + " not possible \n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Construction = new Construction();
		// Test  A
		// Number : 111
		// K      :  1
		task.constructWithoutK(111, 1);
		// Test  B
		// Number :  1
		// K      :  1
		task.constructWithoutK(1, 1);
		// Test  C
		// Number :  123
		// K      :  2
		task.constructWithoutK(123, 2);
		// Test  D
		// Number :  7331
		// K      :  7
		task.constructWithoutK(7331, 7);
		// Test  E
		// Number :  333
		// K      :  3
		task.constructWithoutK(333, 3);
	}
}

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)
// Kotlin program for
// Find two numbers whose sum is N 
// and does not contain any digit as K
class Construction
{
	fun checkDigitK(number: Int, k: Int): Boolean
	{
		var temp: Int = number;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				// When number contains k digit
				return true;
			}
			temp = temp / 10;
		}
		return false;
	}
	fun constructWithoutK(n: Int, k: Int): Unit
	{
		if (n <= 0 || (k < 0 || k > 9))
		{
			return;
		}
		var temp: Int = n;
		var extra: Int = 0;
		var count: Int = 1;
		while (temp != 0)
		{
			if (temp % 10 == k)
			{
				extra += count;
			}
			temp = temp / 10;
			count = count * 10;
		}
		print("\n Given sum : " + n);
		print("\n Given k : " + k);
		if (extra == 0 && !this.checkDigitK(n, k))
		{
			print("\n " + n + " (" + 0 + " + " + n + ") \n");
			return;
		}
		temp = n - extra;
		if (extra > temp)
		{
			// Swap value
			count = extra;
			extra = temp;
			temp = count;
		}
		while (extra < temp)
		{
			if (!this.checkDigitK(temp, k) && 
                !this.checkDigitK(extra, k))
			{
				print("\n " + n + " (" + temp + " + " + extra + ") \n");
				return;
			}
			extra += 1;
			temp -= 1;
		}
		print("\n number : Without digit " + n + 
              " value " + k + " not possible \n");
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Construction = Construction();
	// Test  A
	// Number : 111
	// K      :  1
	task.constructWithoutK(111, 1);
	// Test  B
	// Number :  1
	// K      :  1
	task.constructWithoutK(1, 1);
	// Test  C
	// Number :  123
	// K      :  2
	task.constructWithoutK(123, 2);
	// Test  D
	// Number :  7331
	// K      :  7
	task.constructWithoutK(7331, 7);
	// Test  E
	// Number :  333
	// K      :  3
	task.constructWithoutK(333, 3);
}

Output

 Given sum : 111
 Given k : 1
 111 (89 + 22)

 Given sum : 1
 Given k : 1
 number : Without digit 1 value 1 not possible

 Given sum : 123
 Given k : 2
 123 (113 + 10)

 Given sum : 7331
 Given k : 7
 7331 (6331 + 1000)

 Given sum : 333
 Given k : 3
 333 (222 + 111)

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