Find the frequency of a digit in a number

Given a number N and a digit D. Write a program to find how many times the digit D appears in the number N.

Examples :

Input: N = 1122322  ,  D = 2
Output: 4

Input: N = 346488  ,  D = 9
Output: 0

The idea to solve this problem is to keep extracting digits from the number N and check the extracted digits with the given digit D. If the extracted digit is equals to the digit D then increment the count. 

Below is the implementation of above approach. 

C++

// C++ program to find the frequency
// of a digit in a number
#include <bits/stdc++.h>
using namespace std;

// function to find frequency of digit
// in a number
int frequencyDigits(int n, int d)
{
    // Counter variable to store
    // the frequency
    int c = 0;

    // iterate till number reduces to zero
    while (n > 0) {

        // check for equality
        if (n % 10 == d)
            c++;
        // reduce the number
        n = n / 10;
    }

    return c;
}

// Driver Code
int main()
{

    // input number N
    int N = 1122322;

    // input digit D
    int D = 2;

    cout<<frequencyDigits(N,D);

    return 0;
}

Java

// Java program to find
// the frequency of a
// digit in a number
class GFG
{

// function to find frequency
// of digit in a number
static int frequencyDigits(int n,
                           int d)
{
    // Counter variable to
    // store the frequency
    int c = 0;

    // iterate till number
    // reduces to zero
    while (n > 0)
    {

        // check for equality
        if (n % 10 == d)
            c++;
        // reduce the number
        n = n / 10;
    }
    return c;
}

// Driver Code
public static void main(String args[])
{

    // input number N
    int N = 1122322;

    // input digit D
    int D = 2;

    System.out.println(frequencyDigits(N, D));
}
}

// This code is contributed by Arnab Kundu

Python3

# Python3 program to find the
# frequency of a digit in a number

# function to find frequency
# of digit in a number
def frequencyDigits(n, d):

    # Counter variable to
    # store the frequency
    c = 0;

    # iterate till number
    # reduces to zero
    while (n > 0):

        # check for equality
        if (n % 10 == d):
            c += 1;
        # reduce the number
        n = int(n / 10);

    return c;

# Driver Code

# input number N
N = 1122322;

# input digit D
D = 2;

print(frequencyDigits(N, D));

# This code is contributed by mits

C#

// C# program to find the frequency
// of a digit in a number
using System;

class GFG
{

// function to find frequency
// of digit in a number
static int frequencyDigits(int n,
                           int d)
{
    // Counter variable to
    // store the frequency
    int c = 0;

    // iterate till number
    // reduces to zero
    while (n > 0)
    {

        // check for equality
        if (n % 10 == d)
            c++;
        // reduce the number
        n = n / 10;
    }
    return c;
}

// Driver Code
static public void Main(String []args)
{

    // input number N
    int N = 1122322;

    // input digit D
    int D = 2;

    Console.WriteLine(frequencyDigits(N, D));

}
}

// This code is contributed by Arnab Kundu

PHP

<?php
// PHP program to find the frequency
// of a digit in a number

// function to find frequency
// of digit  in a number
function frequencyDigits($n, $d)
{
    // Counter variable to
    // store the frequency
    $c = 0;

    // iterate till number
    // reduces to zero
    while ($n > 0)
    {

        // check for equality
        if ($n % 10 == $d)
            $c++;
        // reduce the number
        $n = $n / 10;
    }

    return $c;
}

// Driver Code

// input number N
$N = 1122322;

// input digit D
$D = 2;

echo frequencyDigits($N, $D);

// This code is contributed by mits
?>

Javascript

<script>

// Javascript program to find
// the frequency of a
// digit in a number

// Function to find frequency
// of digit in a number
function frequencyDigits(n, d)
{

    // Counter variable to
    // store the frequency
    var c = 0;

    // Iterate till number
    // reduces to zero
    while (n > 0)
    {

        // Check for equality
        if (n % 10 == d)
            c++;

        // Reduce the number
        n = parseInt(n / 10);
    }
    return c;
}

// Driver code

// input number N
var N = 1122322;

// input digit D
var D = 2;

document.write(frequencyDigits(N, D));

// This code is contributed by Khushboogoyal499

</script>

Output :

4

Time Complexity: O(log₁₀N)
Auxiliary Space: O(1)