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# Occurrence of a digit in a given number using C++

## Number of times the digit x occurs in the given number

In this C++ program we will count the number of occurrences of a given digit in the given input number.
The input may lie within the range of integer.

If the digit does not occur in the input it should print 0 else the count of digits.

Sample Input :

Enter a number : 897982
Enter the digit : 9

Output : 2

Explanation : The digit 9 occurs twice

## Algorithm

1.  Start
2. Get the input value from the user.
3. Get the digit from the i/o console.
4. Declare variables n,d,count
• n – Given number
• d – Digit
• count – no. of occurrences
5. Take a while loop.
6. Declare a variable k to store every digit of the number to be compared.
7. Compare k with the digit
• if k equals digit increment count.
8. n=n/10
9. Print the value of count.
10. End

## C++ code

#include <bits/stdc++.h>

using namespace std;

int main()
{
//Declare variables

int n; //given integer
int d; //given digit
int count=0; //declare counter variable

cin >> n >> d; // take input from user

while(n
{

int k = n%10; // to store the digits of the given input

n=n/10

if(k==d) // compare the given digit with digit of input
{
count++; // increment counter variable
}

}

cout << count; // display count of digits

return 0;
}

### One comment on “Occurrence of a digit in a given number using C++”

• Shubham

public class JavaApplication1
{
public static void main(String[] args)
{
Scanner obj = new Scanner(System.in);
int x,c=0;
x = obj.nextInt();
char b = obj.next().charAt(0);
String s = Integer.toString(x);
for(int i=0;i<s.length();i++)
{
if(s.charAt(i)==b)
c=c+1;
}
System.out.println(c);
}
}