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);
    }
    }