Number of common digits present in two given numbers

Here given code implementation process.

// C Program for
// Number of common digits present in two given numbers
#include <stdio.h>
int absValue(int num)
{
	if (num < 0)
	{
		return -num;
	}
	return num;
}
void countCommonDigit(int a, int b)
{
	int n = absValue(a);
	int count = 0;
	// Use to collect digit frequency
	int digit[10];
	// Set initial frequency of digit 0-9
	for (int i = 0; i < 10; ++i)
	{
		digit[i] = 0;
	}
	// If digits are present in A, then set its frequency to 1.
	while (n > 0)
	{
		digit[n % 10] = 1;
		// Remove last digit
		n = n / 10;
	}
	n = absValue(b);
	// Set the normal digit frequency to 2 when 
	// numbers are present in A and B.
	while (n > 0)
	{
		if (digit[n % 10] == 1)
		{
			digit[n % 10] = 2;
		}
		// Remove last digit
		n = n / 10;
	}
	printf("\n Given a : %d  b : %d", a, b);
	// Count common digit
	for (int i = 0; i < 10; ++i)
	{
		if (digit[i] == 2)
		{
			count += 1;
		}
	}
	// Display calculated result
	printf("\n Result : %d\n", count);
}
int main(int argc, char
	const *argv[])
{
	// Test A
	// a = 1241363  b = 523453
	// common = [2 3 4]
	// Output = 3
	countCommonDigit(1241363, 523453);
	// Test B
	// a = 421870  b = 59137
	// common = [1 7] 
	// Output = 2
	countCommonDigit(421870, 59137);
	// Test C
	// a = 534512  b = -132352
	// common = [1 2 3 5]
	// Output = 4
	countCommonDigit(534512, -132352);
	return 0;
}

Output

 Given a : 1241363  b : 523453
 Result : 3

 Given a : 421870  b : 59137
 Result : 2

 Given a : 534512  b : -132352
 Result : 4
// Java program for
// Number of common digits present in two given numbers
public class Counting
{
	public int absValue(int num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	public void countCommonDigit(int a, int b)
	{
		int n = absValue(a);
		int count = 0;
		// Use to collect digit frequency
		int[] digit = new int[10];
		// Set initial frequency of digit 0-9
		for (int i = 0; i < 10; ++i)
		{
			digit[i] = 0;
		}
		// If digits are present in A, then set its frequency to 1.
		while (n > 0)
		{
			digit[n % 10] = 1;
			// Remove last digit
			n = n / 10;
		}
		n = absValue(b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while (n > 0)
		{
			if (digit[n % 10] == 1)
			{
				digit[n % 10] = 2;
			}
			// Remove last digit
			n = n / 10;
		}
		System.out.print("\n Given a : " + a + " b : " + b);
		// Count common digit
		for (int i = 0; i < 10; ++i)
		{
			if (digit[i] == 2)
			{
				count += 1;
			}
		}
		// Display calculated result
		System.out.print("\n Result : " + count + "\n");
	}
	public static void main(String[] args)
	{
		Counting task = new Counting();
		// Test A
		// a = 1241363  b = 523453
		// common = [2 3 4]
		// Output = 3
		task.countCommonDigit(1241363, 523453);
		// Test B
		// a = 421870  b = 59137
		// common = [1 7] 
		// Output = 2
		task.countCommonDigit(421870, 59137);
		// Test C
		// a = 534512  b = -132352
		// common = [1 2 3 5]
		// Output = 4
		task.countCommonDigit(534512, -132352);
	}
}

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
// Include header file
#include <iostream>
using namespace std;

// C++ program for
// Number of common digits present in two given numbers

class Counting
{
	public: int absValue(int num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	void countCommonDigit(int a, int b)
	{
		int n = this->absValue(a);
		int count = 0;
		// Use to collect digit frequency
		int digit[10];
		// Set initial frequency of digit 0-9
		for (int i = 0; i < 10; ++i)
		{
			digit[i] = 0;
		}
		// If digits are present in A, then set its frequency to 1.
		while (n > 0)
		{
			digit[n % 10] = 1;
			// Remove last digit
			n = n / 10;
		}
		n = this->absValue(b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while (n > 0)
		{
			if (digit[n % 10] == 1)
			{
				digit[n % 10] = 2;
			}
			// Remove last digit
			n = n / 10;
		}
		cout << "\n Given a : " << a << " b : " << b;
		// Count common digit
		for (int i = 0; i < 10; ++i)
		{
			if (digit[i] == 2)
			{
				count += 1;
			}
		}
		// Display calculated result
		cout << "\n Result : " << count << "\n";
	}
};
int main()
{
	Counting *task = new Counting();
	// Test A
	// a = 1241363  b = 523453
	// common = [2 3 4]
	// Output = 3
	task->countCommonDigit(1241363, 523453);
	// Test B
	// a = 421870  b = 59137
	// common = [1 7] 
	// Output = 2
	task->countCommonDigit(421870, 59137);
	// Test C
	// a = 534512  b = -132352
	// common = [1 2 3 5]
	// Output = 4
	task->countCommonDigit(534512, -132352);
	return 0;
}

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
package main
import "fmt"
// Go program for
// Number of common digits present in two given numbers
type Counting struct {}
func getCounting() * Counting {
	var me *Counting = &Counting {}
	return me
}
func(this Counting) absValue(num int) int {
	if num < 0 {
		return -num
	}
	return num
}
func(this Counting) countCommonDigit(a, b int) {
	var n int = this.absValue(a)
	var count int = 0
	// Use to collect digit frequency
	var digit = make([] int, 10)
	// Set initial frequency of digit 0-9
	for i := 0 ; i < 10 ; i++ {
		digit[i] = 0
	}
	// If digits are present in A, then set its frequency to 1.
	for (n > 0) {
		digit[n % 10] = 1
		// Remove last digit
		n = n / 10
	}
	n = this.absValue(b)
	// Set the normal digit frequency to 2 when 
	// numbers are present in A and B.
	for (n > 0) {
		if digit[n % 10] == 1 {
			digit[n % 10] = 2
		}
		// Remove last digit
		n = n / 10
	}
	fmt.Print("\n Given a : ", a, " b : ", b)
	// Count common digit
	for i := 0 ; i < 10 ; i++ {
		if digit[i] == 2 {
			count += 1
		}
	}
	// Display calculated result
	fmt.Print("\n Result : ", count, "\n")
}
func main() {
	var task * Counting = getCounting()
	// Test A
	// a = 1241363  b = 523453
	// common = [2 3 4]
	// Output = 3
	task.countCommonDigit(1241363, 523453)
	// Test B
	// a = 421870  b = 59137
	// common = [1 7] 
	// Output = 2
	task.countCommonDigit(421870, 59137)
	// Test C
	// a = 534512  b = -132352
	// common = [1 2 3 5]
	// Output = 4
	task.countCommonDigit(534512, -132352)
}

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
// Include namespace system
using System;
// Csharp program for
// Number of common digits present in two given numbers
public class Counting
{
	public int absValue(int num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	public void countCommonDigit(int a, int b)
	{
		int n = this.absValue(a);
		int count = 0;
		// Use to collect digit frequency
		int[] digit = new int[10];
		// Set initial frequency of digit 0-9
		for (int i = 0; i < 10; ++i)
		{
			digit[i] = 0;
		}
		// If digits are present in A, then set its frequency to 1.
		while (n > 0)
		{
			digit[n % 10] = 1;
			// Remove last digit
			n = n / 10;
		}
		n = this.absValue(b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while (n > 0)
		{
			if (digit[n % 10] == 1)
			{
				digit[n % 10] = 2;
			}
			// Remove last digit
			n = n / 10;
		}
		Console.Write("\n Given a : " + a + " b : " + b);
		// Count common digit
		for (int i = 0; i < 10; ++i)
		{
			if (digit[i] == 2)
			{
				count += 1;
			}
		}
		// Display calculated result
		Console.Write("\n Result : " + count + "\n");
	}
	public static void Main(String[] args)
	{
		Counting task = new Counting();
		// Test A
		// a = 1241363  b = 523453
		// common = [2 3 4]
		// Output = 3
		task.countCommonDigit(1241363, 523453);
		// Test B
		// a = 421870  b = 59137
		// common = [1 7] 
		// Output = 2
		task.countCommonDigit(421870, 59137);
		// Test C
		// a = 534512  b = -132352
		// common = [1 2 3 5]
		// Output = 4
		task.countCommonDigit(534512, -132352);
	}
}

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
<?php
// Php program for
// Number of common digits present in two given numbers
class Counting
{
	public	function absValue($num)
	{
		if ($num < 0)
		{
			return -$num;
		}
		return $num;
	}
	public	function countCommonDigit($a, $b)
	{
		$n = $this->absValue($a);
		$count = 0;
		// Use to collect digit frequency
		$digit = array_fill(0, 10, 0);
		// If digits are present in A, then set its frequency to 1.
		while ($n > 0)
		{
			$digit[$n % 10] = 1;
			// Remove last digit
			$n = (int)($n / 10);
		}
		$n = $this->absValue($b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while ($n > 0)
		{
			if ($digit[$n % 10] == 1)
			{
				$digit[$n % 10] = 2;
			}
			// Remove last digit
			$n = (int)($n / 10);
		}
		echo("\n Given a : ".$a.
			" b : ".$b);
		// Count common digit
		for ($i = 0; $i < 10; ++$i)
		{
			if ($digit[$i] == 2)
			{
				$count += 1;
			}
		}
		// Display calculated result
		echo("\n Result : ".$count."\n");
	}
}

function main()
{
	$task = new Counting();
	// Test A
	// a = 1241363  b = 523453
	// common = [2 3 4]
	// Output = 3
	$task->countCommonDigit(1241363, 523453);
	// Test B
	// a = 421870  b = 59137
	// common = [1 7] 
	// Output = 2
	$task->countCommonDigit(421870, 59137);
	// Test C
	// a = 534512  b = -132352
	// common = [1 2 3 5]
	// Output = 4
	$task->countCommonDigit(534512, -132352);
}
main();

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
// Node JS program for
// Number of common digits present in two given numbers
class Counting
{
	absValue(num)
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	countCommonDigit(a, b)
	{
		var n = this.absValue(a);
		var count = 0;
		// Use to collect digit frequency
		var digit = Array(10).fill(0);
		// If digits are present in A, then set its frequency to 1.
		while (n > 0)
		{
			digit[n % 10] = 1;
			// Remove last digit
			n = parseInt(n / 10);
		}
		n = this.absValue(b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while (n > 0)
		{
			if (digit[n % 10] == 1)
			{
				digit[n % 10] = 2;
			}
			// Remove last digit
			n = parseInt(n / 10);
		}
		process.stdout.write("\n Given a : " + a + " b : " + b);
		// Count common digit
		for (var i = 0; i < 10; ++i)
		{
			if (digit[i] == 2)
			{
				count += 1;
			}
		}
		// Display calculated result
		process.stdout.write("\n Result : " + count + "\n");
	}
}

function main()
{
	var task = new Counting();
	// Test A
	// a = 1241363  b = 523453
	// common = [2 3 4]
	// Output = 3
	task.countCommonDigit(1241363, 523453);
	// Test B
	// a = 421870  b = 59137
	// common = [1 7] 
	// Output = 2
	task.countCommonDigit(421870, 59137);
	// Test C
	// a = 534512  b = -132352
	// common = [1 2 3 5]
	// Output = 4
	task.countCommonDigit(534512, -132352);
}
main();

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
#  Python 3 program for
#  Number of common digits present in two given numbers
class Counting :
	def absValue(self, num) :
		if (num < 0) :
			return -num
		
		return num
	
	def countCommonDigit(self, a, b) :
		n = self.absValue(a)
		count = 0
		#  Use to collect digit frequency
		digit = [0] * (10)
		#  If digits are present in A, then set its frequency to 1.
		while (n > 0) :
			digit[n % 10] = 1
			#  Remove last digit
			n = int(n / 10)
		
		n = self.absValue(b)
		#  Set the normal digit frequency to 2 when 
		#  numbers are present in A and B.
		while (n > 0) :
			if (digit[n % 10] == 1) :
				digit[n % 10] = 2
			
			#  Remove last digit
			n = int(n / 10)
		
		print("\n Given a : ", a ," b : ", b, end = "")
		i = 0
		#  Count common digit
		while (i < 10) :
			if (digit[i] == 2) :
				count += 1
			
			i += 1
		
		#  Display calculated result
		print("\n Result : ", count )
	

def main() :
	task = Counting()
	#  Test A
	#  a = 1241363  b = 523453
	#  common = [2 3 4]
	#  Output = 3
	task.countCommonDigit(1241363, 523453)
	#  Test B
	#  a = 421870  b = 59137
	#  common = [1 7] 
	#  Output = 2
	task.countCommonDigit(421870, 59137)
	#  Test C
	#  a = 534512  b = -132352
	#  common = [1 2 3 5]
	#  Output = 4
	task.countCommonDigit(534512, -132352)

if __name__ == "__main__": main()

Output

 Given a :  1241363  b :  523453
 Result :  3

 Given a :  421870  b :  59137
 Result :  2

 Given a :  534512  b :  -132352
 Result :  4
#  Ruby program for
#  Number of common digits present in two given numbers
class Counting 
	def absValue(num) 
		if (num < 0) 
			return -num
		end

		return num
	end

	def countCommonDigit(a, b) 
		n = self.absValue(a)
		count = 0
		#  Use to collect digit frequency
		digit = Array.new(10) {0}
		#  If digits are present in A, then set its frequency to 1.
		while (n > 0) 
			digit[n % 10] = 1
			#  Remove last digit
			n = n / 10
		end

		n = self.absValue(b)
		#  Set the normal digit frequency to 2 when 
		#  numbers are present in A and B.
		while (n > 0) 
			if (digit[n % 10] == 1) 
				digit[n % 10] = 2
			end

			#  Remove last digit
			n = n / 10
		end

		print("\n Given a : ", a ," b : ", b)
		i = 0
		#  Count common digit
		while (i < 10) 
			if (digit[i] == 2) 
				count += 1
			end

			i += 1
		end

		#  Display calculated result
		print("\n Result : ", count ,"\n")
	end

end

def main() 
	task = Counting.new()
	#  Test A
	#  a = 1241363  b = 523453
	#  common = [2 3 4]
	#  Output = 3
	task.countCommonDigit(1241363, 523453)
	#  Test B
	#  a = 421870  b = 59137
	#  common = [1 7] 
	#  Output = 2
	task.countCommonDigit(421870, 59137)
	#  Test C
	#  a = 534512  b = -132352
	#  common = [1 2 3 5]
	#  Output = 4
	task.countCommonDigit(534512, -132352)
end

main()

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
// Scala program for
// Number of common digits present in two given numbers
class Counting()
{
	def absValue(num: Int): Int = {
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	def countCommonDigit(a: Int, b: Int): Unit = {
		var n: Int = absValue(a);
		var count: Int = 0;
		// Use to collect digit frequency
		var digit: Array[Int] = Array.fill[Int](10)(0);
		// If digits are present in A, then set its frequency to 1.
		while (n > 0)
		{
			digit(n % 10) = 1;
			// Remove last digit
			n = n / 10;
		}
		n = absValue(b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while (n > 0)
		{
			if (digit(n % 10) == 1)
			{
				digit(n % 10) = 2;
			}
			// Remove last digit
			n = n / 10;
		}
		print("\n Given a : " + a + " b : " + b);
		var i: Int = 0;
		// Count common digit
		while (i < 10)
		{
			if (digit(i) == 2)
			{
				count += 1;
			}
			i += 1;
		}
		// Display calculated result
		print("\n Result : " + count + "\n");
	}
}
object Main
{
	def main(args: Array[String]): Unit = {
		var task: Counting = new Counting();
		// Test A
		// a = 1241363  b = 523453
		// common = [2 3 4]
		// Output = 3
		task.countCommonDigit(1241363, 523453);
		// Test B
		// a = 421870  b = 59137
		// common = [1 7] 
		// Output = 2
		task.countCommonDigit(421870, 59137);
		// Test C
		// a = 534512  b = -132352
		// common = [1 2 3 5]
		// Output = 4
		task.countCommonDigit(534512, -132352);
	}
}

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4
// Swift 4 program for
// Number of common digits present in two given numbers
class Counting
{
	func absValue(_ num: Int) -> Int
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	func countCommonDigit(_ a: Int, _ b: Int)
	{
		var n: Int = self.absValue(a);
		var count: Int = 0;
		// Use to collect digit frequency
		var digit: [Int] = Array(repeating: 0, count: 10);
		// If digits are present in A, then set its frequency to 1.
		while (n > 0)
		{
			digit[n % 10] = 1;
			// Remove last digit
			n = n / 10;
		}
		n = self.absValue(b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while (n > 0)
		{
			if (digit[n % 10] == 1)
			{
				digit[n % 10] = 2;
			}
			// Remove last digit
			n = n / 10;
		}
		print("\n Given a : ", a ," b : ", b, terminator: "");
		var i: Int = 0;
		// Count common digit
		while (i < 10)
		{
			if (digit[i] == 2)
			{
				count += 1;
			}
			i += 1;
		}
		// Display calculated result
		print("\n Result : ", count );
	}
}
func main()
{
	let task: Counting = Counting();
	// Test A
	// a = 1241363  b = 523453
	// common = [2 3 4]
	// Output = 3
	task.countCommonDigit(1241363, 523453);
	// Test B
	// a = 421870  b = 59137
	// common = [1 7] 
	// Output = 2
	task.countCommonDigit(421870, 59137);
	// Test C
	// a = 534512  b = -132352
	// common = [1 2 3 5]
	// Output = 4
	task.countCommonDigit(534512, -132352);
}
main();

Output

 Given a :  1241363  b :  523453
 Result :  3

 Given a :  421870  b :  59137
 Result :  2

 Given a :  534512  b :  -132352
 Result :  4
// Kotlin program for
// Number of common digits present in two given numbers
class Counting
{
	fun absValue(num: Int): Int
	{
		if (num < 0)
		{
			return -num;
		}
		return num;
	}
	fun countCommonDigit(a: Int, b: Int): Unit
	{
		var n: Int = this.absValue(a);
		var count: Int = 0;
		// Use to collect digit frequency
		var digit: Array < Int > = Array(10)
		{
			0
		};
		// If digits are present in A, then set its frequency to 1.
		while (n > 0)
		{
			digit[n % 10] = 1;
			// Remove last digit
			n = n / 10;
		}
		n = this.absValue(b);
		// Set the normal digit frequency to 2 when 
		// numbers are present in A and B.
		while (n > 0)
		{
			if (digit[n % 10] == 1)
			{
				digit[n % 10] = 2;
			}
			// Remove last digit
			n = n / 10;
		}
		print("\n Given a : " + a + " b : " + b);
		var i: Int = 0;
		// Count common digit
		while (i < 10)
		{
			if (digit[i] == 2)
			{
				count += 1;
			}
			i += 1;
		}
		// Display calculated result
		print("\n Result : " + count + "\n");
	}
}
fun main(args: Array < String > ): Unit
{
	val task: Counting = Counting();
	// Test A
	// a = 1241363  b = 523453
	// common = [2 3 4]
	// Output = 3
	task.countCommonDigit(1241363, 523453);
	// Test B
	// a = 421870  b = 59137
	// common = [1 7] 
	// Output = 2
	task.countCommonDigit(421870, 59137);
	// Test C
	// a = 534512  b = -132352
	// common = [1 2 3 5]
	// Output = 4
	task.countCommonDigit(534512, -132352);
}

Output

 Given a : 1241363 b : 523453
 Result : 3

 Given a : 421870 b : 59137
 Result : 2

 Given a : 534512 b : -132352
 Result : 4


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