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 Nint 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 Codeint 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 Recursionimport 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 Ndef 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 Codeif __name__ == "__main__": n = 1000; k = 0; print(countdigits(n, k)) |
C#
// C# implementation to count// the occurrence of a digit in// number using Recursionusing 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 Nfunction 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 Codevar n = 1000;var k = 0;document.write( countdigits(n, k));</script> |
3
Attention reader! Don’t stop learning now. Get hold of all the important mathematical concepts for competitive programming with the Essential Maths for CP Course at a student-friendly price. To complete your preparation from learning a language to DS Algo and many more, please refer Complete Interview Preparation Course.
