Given an integer N, count and return the number of zeros that are present in the given integer using recursion.
Count Zeros
PROBLEM:-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 :
4SOLUTION:-
#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;
}