Given an integer N, count and return the number of zeros that are present in the given integer using recursion.

PROBLEM:-

Count Zeros

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Given an integer N, count and return the number of zeros that are present in the given integer using recursion.

Input Format :

Integer N

Output Format :

Number of zeros in N

Constraints :

0 <= N <= 10^9

Sample Input 1 :

10204

Sample Output 1 :

2

Sample Input 2 :

708000

Sample Output 2 :

4

SOLUTION:-

#include <iostream>
using namespace std;

int countZeros(int n) 
{
    // Write your code here
   
   if(n==0)
   return 0;
    else{
   int d=n%10;
   if(d==0)
   {
       return countZeros(n/10)+1;
   }
   else{
       countZeros(n/10);
   }
    }
       
        
}

int main() {
    int n;
    cin >> n;
    cout << countZeros(n) << endl;
}




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