# Print all distinct uncommon digits present in two given numbers

Here given code implementation process.

``````// C Program for
// Print all distinct uncommon digits present in two given numbers
#include <stdio.h>

int absValue(int num)
{
if (num < 0)
{
return -num;
}
return num;
}
void unCommonDigit(int a, int b)
{
int n = absValue(a);
int count = 0;
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);
while (n > 0)
{
if (digit[n % 10] == 0)
{
// new digit in B number
digit[n % 10] = 3;
}
else if (digit[n % 10] == 1)
{
// Common digit
digit[n % 10] = 2;
}
// Remove last digit
n = n / 10;
}
printf("\n Given a : %d  b : %d\n", a, b);
for (int i = 0; i < 10; ++i)
{
if (digit[i] == 1 || digit[i] == 3)
{
count++;
printf(" %d", i);
}
}
if (count == 0)
{
printf("\n None \n");
}
}
int main(int argc, char
const *argv[])
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
unCommonDigit(1241363, 523753);
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
unCommonDigit(421870, 59137);
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
unCommonDigit(534512, -137352);
return 0;
}``````

#### Output

`````` Given a : 1241363  b : 523753
1 4 5 6 7
Given a : 421870  b : 59137
0 2 3 4 5 8 9
Given a : 534512  b : -137352
4 7``````
``````// Java program for
// Print all distinct uncommon digits present in two given numbers
public class UncommonDigits
{
public int absValue(int num)
{
if (num < 0)
{
return -num;
}
return num;
}
public void unCommonDigit(int a, int b)
{
int n = absValue(a);
boolean status = false;
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);
while (n > 0)
{
if (digit[n % 10] == 0)
{
// new digit in B number
digit[n % 10] = 3;
}
else if (digit[n % 10] == 1)
{
// Common digit
digit[n % 10] = 2;
}
// Remove last digit
n = n / 10;
}
System.out.print("\n Given a : " + a + " b : " + b + "\n");
for (int i = 0; i < 10; ++i)
{
if (digit[i] == 1 || digit[i] == 3)
{
status = true;
System.out.print(" " + i);
}
}
if (status == false)
{
System.out.print("\n None \n");
}
}
public static void main(String[] args)
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}
}``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````// Include header file
#include <iostream>
using namespace std;
// C++ program for
// Print all distinct uncommon digits present in two given numbers
class UncommonDigits
{
public: int absValue(int num)
{
if (num < 0)
{
return -num;
}
return num;
}
void unCommonDigit(int a, int b)
{
int n = this->absValue(a);
bool status = false;
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);
while (n > 0)
{
if (digit[n % 10] == 0)
{
// new digit in B number
digit[n % 10] = 3;
}
else if (digit[n % 10] == 1)
{
// Common digit
digit[n % 10] = 2;
}
// Remove last digit
n = n / 10;
}
cout << "\n Given a : " << a << " b : " << b << "\n";
for (int i = 0; i < 10; ++i)
{
if (digit[i] == 1 || digit[i] == 3)
{
status = true;
cout << " " << i;
}
}
if (status == false)
{
cout << "\n None \n";
}
}
};
int main()
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
return 0;
}``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````package main
import "fmt"
// Go program for
// Print all distinct uncommon digits present in two given numbers
type UncommonDigits struct {}
func getUncommonDigits() * UncommonDigits {
var me *UncommonDigits = &UncommonDigits {}
return me
}
func(this UncommonDigits) absValue(num int) int {
if num < 0 {
return -num
}
return num
}
func(this UncommonDigits) unCommonDigit(a, b int) {
var n int = this.absValue(a)
var status bool = false
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)
for (n > 0) {
if digit[n % 10] == 0 {
// new digit in B number
digit[n % 10] = 3
} else if digit[n % 10] == 1 {
// Common digit
digit[n % 10] = 2
}
// Remove last digit
n = n / 10
}
fmt.Print("\n Given a : ", a, " b : ", b, "\n")
for i := 0 ; i < 10 ; i++ {
if digit[i] == 1 || digit[i] == 3 {
status = true
fmt.Print(" ", i)
}
}
if status == false {
fmt.Print("\n None \n")
}
}
func main() {
var task * UncommonDigits = getUncommonDigits()
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````// Include namespace system
using System;
// Csharp program for
// Print all distinct uncommon digits present in two given numbers
public class UncommonDigits
{
public int absValue(int num)
{
if (num < 0)
{
return -num;
}
return num;
}
public void unCommonDigit(int a, int b)
{
int n = this.absValue(a);
Boolean status = false;
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);
while (n > 0)
{
if (digit[n % 10] == 0)
{
// new digit in B number
digit[n % 10] = 3;
}
else if (digit[n % 10] == 1)
{
// Common digit
digit[n % 10] = 2;
}
// Remove last digit
n = n / 10;
}
Console.Write("\n Given a : " + a + " b : " + b + "\n");
for (int i = 0; i < 10; ++i)
{
if (digit[i] == 1 || digit[i] == 3)
{
status = true;
Console.Write(" " + i);
}
}
if (status == false)
{
Console.Write("\n None \n");
}
}
public static void Main(String[] args)
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}
}``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````<?php
// Php program for
// Print all distinct uncommon digits present in two given numbers
class UncommonDigits
{
public	function absValue(\$num)
{
if (\$num < 0)
{
return -\$num;
}
return \$num;
}
public	function unCommonDigit(\$a, \$b)
{
\$n = \$this->absValue(\$a);
\$status = false;
\$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);
while (\$n > 0)
{
if (\$digit[\$n % 10] == 0)
{
// new digit in B number
\$digit[\$n % 10] = 3;
}
else if (\$digit[\$n % 10] == 1)
{
// Common digit
\$digit[\$n % 10] = 2;
}
// Remove last digit
\$n = (int)(\$n / 10);
}
echo("\n Given a : ".\$a.
" b : ".\$b.
"\n");
for (\$i = 0; \$i < 10; ++\$i)
{
if (\$digit[\$i] == 1 || \$digit[\$i] == 3)
{
\$status = true;
echo(" ".\$i);
}
}
if (\$status == false)
{
echo("\n None \n");
}
}
}

function main()
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}
main();``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````// Node JS program for
// Print all distinct uncommon digits present in two given numbers
class UncommonDigits
{
absValue(num)
{
if (num < 0)
{
return -num;
}
return num;
}
unCommonDigit(a, b)
{
var n = this.absValue(a);
var status = false;
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);
while (n > 0)
{
if (digit[n % 10] == 0)
{
// new digit in B number
digit[n % 10] = 3;
}
else if (digit[n % 10] == 1)
{
// Common digit
digit[n % 10] = 2;
}
// Remove last digit
n = parseInt(n / 10);
}
process.stdout.write("\n Given a : " + a + " b : " + b + "\n");
for (var i = 0; i < 10; ++i)
{
if (digit[i] == 1 || digit[i] == 3)
{
status = true;
process.stdout.write(" " + i);
}
}
if (status == false)
{
process.stdout.write("\n None \n");
}
}
}

function main()
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}
main();``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````#  Python 3 program for
#  Print all distinct uncommon digits present in two given numbers
class UncommonDigits :
def absValue(self, num) :
if (num < 0) :
return -num

return num

def unCommonDigit(self, a, b) :
n = self.absValue(a)
status = False
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)
while (n > 0) :
if (digit[n % 10] == 0) :
#  new digit in B number
digit[n % 10] = 3
elif (digit[n % 10] == 1) :
#  Common digit
digit[n % 10] = 2

#  Remove last digit
n = int(n / 10)

print("\n Given a : ", a ," b : ", b )
i = 0
while (i < 10) :
if (digit[i] == 1 or digit[i] == 3) :
status = True
print(" ", i, end = "")

i += 1

if (status == False) :
print("\n None ")

def main() :
#  Test A
#  a = 1241363  b = 523753
#  uncommon digit = [1 4 5 6 7]
#  Test B
#  a = 421870  b = 59137
#  uncommon digit = [0 2 3 4 5 8 9]
#  Test C
#  a = 534512  b = -137352
#  uncommon digit = [4 7]

if __name__ == "__main__": main()``````

#### Output

`````` Given a :  1241363  b :  523753
1  4  5  6  7
Given a :  421870  b :  59137
0  2  3  4  5  8  9
Given a :  534512  b :  -137352
4  7``````
``````#  Ruby program for
#  Print all distinct uncommon digits present in two given numbers
class UncommonDigits
def absValue(num)
if (num < 0)
return -num
end

return num
end

def unCommonDigit(a, b)
n = self.absValue(a)
status = false
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)
while (n > 0)
if (digit[n % 10] == 0)
#  new digit in B number
digit[n % 10] = 3
elsif (digit[n % 10] == 1)
#  Common digit
digit[n % 10] = 2
end

#  Remove last digit
n = n / 10
end

print("\n Given a : ", a ," b : ", b ,"\n")
i = 0
while (i < 10)
if (digit[i] == 1 || digit[i] == 3)
status = true
print(" ", i)
end

i += 1
end

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

end

end

def main()
#  Test A
#  a = 1241363  b = 523753
#  uncommon digit = [1 4 5 6 7]
#  Test B
#  a = 421870  b = 59137
#  uncommon digit = [0 2 3 4 5 8 9]
#  Test C
#  a = 534512  b = -137352
#  uncommon digit = [4 7]
end

main()``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````// Scala program for
// Print all distinct uncommon digits present in two given numbers
class UncommonDigits()
{
def absValue(num: Int): Int = {
if (num < 0)
{
return -num;
}
return num;
}
def unCommonDigit(a: Int, b: Int): Unit = {
var n: Int = absValue(a);
var status: Boolean = false;
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);
while (n > 0)
{
if (digit(n % 10) == 0)
{
// new digit in B number
digit(n % 10) = 3;
}
else if (digit(n % 10) == 1)
{
// Common digit
digit(n % 10) = 2;
}
// Remove last digit
n = n / 10;
}
print("\n Given a : " + a + " b : " + b + "\n");
var i: Int = 0;
while (i < 10)
{
if (digit(i) == 1 || digit(i) == 3)
{
status = true;
print(" " + i);
}
i += 1;
}
if (status == false)
{
print("\n None \n");
}
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: UncommonDigits = new UncommonDigits();
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}
}``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````
``````// Swift 4 program for
// Print all distinct uncommon digits present in two given numbers
class UncommonDigits
{
func absValue(_ num: Int) -> Int
{
if (num < 0)
{
return -num;
}
return num;
}
func unCommonDigit(_ a: Int, _ b: Int)
{
var n: Int = self.absValue(a);
var status: Bool = false;
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);
while (n > 0)
{
if (digit[n % 10] == 0)
{
// new digit in B number
digit[n % 10] = 3;
}
else if (digit[n % 10] == 1)
{
// Common digit
digit[n % 10] = 2;
}
// Remove last digit
n = n / 10;
}
print("\n Given a : ", a ," b : ", b );
var i: Int = 0;
while (i < 10)
{
if (digit[i] == 1 || digit[i] == 3)
{
status = true;
print(" ", i, terminator: "");
}
i += 1;
}
if (status == false)
{
print("\n None ");
}
}
}
func main()
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}
main();``````

#### Output

`````` Given a :  1241363  b :  523753
1  4  5  6  7
Given a :  421870  b :  59137
0  2  3  4  5  8  9
Given a :  534512  b :  -137352
4  7``````
``````// Kotlin program for
// Print all distinct uncommon digits present in two given numbers
class UncommonDigits
{
fun absValue(num: Int): Int
{
if (num < 0)
{
return -num;
}
return num;
}
fun unCommonDigit(a: Int, b: Int): Unit
{
var n: Int = this.absValue(a);
var status: Boolean = false;
val 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);
while (n > 0)
{
if (digit[n % 10] == 0)
{
// new digit in B number
digit[n % 10] = 3;
}
else if (digit[n % 10] == 1)
{
// Common digit
digit[n % 10] = 2;
}
// Remove last digit
n = n / 10;
}
print("\n Given a : " + a + " b : " + b + "\n");
var i: Int = 0;
while (i < 10)
{
if (digit[i] == 1 || digit[i] == 3)
{
status = true;
print(" " + i);
}
i += 1;
}
if (status == false)
{
print("\n None \n");
}
}
}
fun main(args: Array < String > ): Unit
{
// Test A
// a = 1241363  b = 523753
// uncommon digit = [1 4 5 6 7]
// Test B
// a = 421870  b = 59137
// uncommon digit = [0 2 3 4 5 8 9]
// Test C
// a = 534512  b = -137352
// uncommon digit = [4 7]
}``````

#### Output

`````` Given a : 1241363 b : 523753
1 4 5 6 7
Given a : 421870 b : 59137
0 2 3 4 5 8 9
Given a : 534512 b : -137352
4 7``````

## Comment

Please share your knowledge to improve code and content standard. Also submit your doubts, and test case. We improve by your feedback. We will try to resolve your query as soon as possible.