Count the occurrence of digit K in a given number N using Recursion

Given an integer N greater than 0, the task is to find the occurrence of digit K present in the given number N.
Examples:

Input: N = 10000, K = 0 
Output: 4 
Explanation:
Occurrence of ‘0’ digit in 10000 is 4.
Input: N = 51435, K = 5 
Output: 2 
Explanation:
Occurrence of ‘5’ digit in 51435 is 2 

Approach: The idea is to use recursion to extract the least significant digit of the number in each step and Check that the digit extracted is equal to the given digit K or not. Finally, recurse for checking the next digits of the number by computing the integer division of that number by 10.
Below is the implementation of the above approach:

C++

// C++ implementation to count
// the occurrence of a digit in
// number using Recursion

#include <bits/stdc++.h>

using namespace std;

// Function to count the digit K
// in the given number N
int countdigits(int n, int k)
{
    if (n == 0)
        return 0;

    // Extracting least
    // significant digit
    int digit = n % 10;
    if (digit == k)
        return 1 + countdigits(n / 10, k);

    return countdigits(n / 10, k);
}

// Driver Code
int main()
{
    int n = 1000;
    int k = 0;

    cout << countdigits(n, k) << endl;
    return 0;
}

Java

// Java implementation to count
// the occurrence of a digit in
// number using Recursion

import java.util.*;

class GFG {

    // Function to count the digit K
    // in the given number N
    static double countdigits(int n, int k)
    {
        if (n == 0)
            return 0;

        // Extracting least
        // significant digit
        int digit = n % 10;
        if (digit == k)
            return 1 + countdigits(n / 10, k);

        return countdigits(n / 10, k);
    }

    // Driver Code
    public static void main(String[] args)
    {
        int n = 1000;
        int k = 0;
        System.out.println(countdigits(n, k));
    }
}

Python3

# Python implementation to count
# the occurrence of a digit in
# number using Recursion

# Function to count the digit K
# in the given number N
def countdigits(n, k):
    if n == 0:
        return 0

    # Extracting least
    # significant digit
    digit = n % 10

    if digit == k:
        return 1 + countdigits(n / 10, k)

    return countdigits(n / 10, k)

# Driver Code
if __name__ == "__main__":
    n = 1000;
    k = 0;
    print(countdigits(n, k))

C#

// C# implementation to count
// the occurrence of a digit in
// number using Recursion

using System;

class GFG {

    // Function to count the digit K
    // in the given number N
    static double countdigits(int n,
                              int k)
    {
        if (n == 0)
            return 0;

        // Extracting least
        // significant digit
        int digit = n % 10;
        if (digit == k)
            return 1 +
             countdigits(n / 10, k);

        return countdigits(n / 10, k);
    }

    // Driver Code
    static public void Main()
    {
        int n = 1000;
        int k = 0;
        Console.WriteLine(countdigits(n, k));
    }
}

Javascript

<script>

// Javascript implementation to count
// the occurrence of a digit in
// number using Recursion


// Function to count the digit K
// in the given number N
function countdigits(n, k)
{
    if (n == 0)
        return 0;

    // Extracting least
    // significant digit
    var digit = n % 10;
    if (digit == k)
        return 1 + countdigits(n / 10, k);

    return countdigits(n / 10, k);
}

// Driver Code
var n = 1000;
var k = 0;
document.write( countdigits(n, k));


</script>

Output:

3

Time Complexity: O(log₁₀n)

Auxiliary Space: O(log₁₀n)