Count numbers having 0 as a digit

Count how many integers from 1 to N contains 0’s as a digit.

Examples:

Input:  n = 9
Output: 0

Input: n = 107
Output: 17
The numbers having 0 are 10, 20,..90, 100, 101..107

Input: n = 155
Output: 24
The numbers having 0 are 10, 20,..90, 100, 101..110,
120, ..150.

The idea is to traverse all numbers from 1 to n. For every traversed number, traverse through its digits, if any digit is 0, increment count. Below is the implementation of the above idea :

C++

// C++ program to count numbers from 1 to n with 
// 0 as a digit 
#include<bits/stdc++.h>  
using namespace std; 
  
// Returns 1 if x has 0, else 0 
int has0(int x) 
{ 
    // Traverse through all digits of 
    // x to check if it has 0. 
    while (x) 
    { 
        // If current digit is 0, return true 
        if (x % 10 == 0) 
          return 1; 
  
        x /= 10; 
    } 
  
    return 0; 
} 
  
// Returns count of numbers from 1 to n with 0 as digit 
int getCount(int n) 
{ 
    // Initialize count of numbers having 0 as digit 
    int count = 0; 
  
    // Traverse through all numbers and for every number 
    // check if it has 0. 
    for (int i=1; i<=n; i++) 
        count += has0(i); 
  
    return count; 
} 
  
// Driver program 
int main() 
{ 
    int n = 107; 
    cout << "Count of numbers from 1" << " to "
         << n << " is " << getCount(n); 
} 

Java

// Java program to count numbers  
// from 1 to n with 0 as a digit 
import java.io.*; 
  
class GFG { 
      
    // Returns 1 if x has 0, else 0 
    static int has0(int x) 
    { 
        // Traverse through all digits 
        // of x to check if it has 0. 
        while (x != 0) 
        { 
            // If current digit is 0, 
            // return true 
            if (x % 10 == 0) 
            return 1; 
      
            x /= 10; 
        } 
      
        return 0; 
    } 
      
    // Returns count of numbers 
    // from 1 to n with 0 as digit 
    static int getCount(int n) 
    { 
        // Initialize count of  
        // numbers having 0 as digit 
        int count = 0; 
      
        // Traverse through all numbers 
        // and for every number 
        // check if it has 0. 
        for (int i = 1; i <= n; i++) 
            count += has0(i); 
      
        return count; 
    } 
  
      
// Driver program 
public static void main(String args[]) 
{ 
  int n = 107; 
  System.out.println("Count of numbers from 1"
            + " to " +n + " is " + getCount(n)); 
} 
} 
  
// This code is contributed by Nikita Tiwari. 

Python3

# Python3 program to count numbers 
# from 1 to n with 0 as a digit 
  
# Returns 1 if x has 0, else 0 
def has0(x) : 
      
    # Traverse through all digits  
    # of x to check if it has 0. 
    while (x != 0) : 
          
        # If current digit is 0, 
        # return true 
        if (x % 10 == 0) : 
            return 1
  
        x = x // 10
      
    return 0
  
  
# Returns count of numbers  
# from 1 to n with 0 as digit 
def getCount(n) : 
      
    # Initialize count of numbers 
    # having 0 as digit. 
    count = 0
  
    # Traverse through all numbers 
    # and for every number check  
    # if it has 0. 
    for i in range(1, n + 1) : 
        count = count + has0(i) 
  
    return count 
  
  
# Driver program 
n = 107
print("Count of numbers from 1", " to ", 
                n , " is " , getCount(n)) 
  
  
# This code is contributed by Nikita tiwari. 

C#

// C# program to count numbers  
// from 1 to n with 0 as a digit 
using System; 
  
class GFG 
{ 
      
    // Returns 1 if x has 0, else 0 
    static int has0(int x) 
    { 
        // Traverse through all digits 
        // of x to check if it has 0. 
        while (x != 0) 
        { 
            // If current digit is 0, 
            // return true 
            if (x % 10 == 0) 
            return 1; 
      
            x /= 10; 
        } 
      
        return 0; 
    } 
      
    // Returns count of numbers 
    // from 1 to n with 0 as digit 
    static int getCount(int n) 
    { 
        // Initialize count of  
        // numbers having 0 as digit 
        int count = 0; 
      
        // Traverse through all numbers 
        // and for every number 
        // check if it has 0. 
        for (int i = 1; i <= n; i++) 
            count += has0(i); 
      
        return count; 
    } 
  
      
// Driver Code 
public static void Main() 
{ 
      
    int n = 107; 
    Console.WriteLine("Count of numbers from 1"
                        + " to " +n + " is " + getCount(n)); 
} 
} 
  
// This code is contributed by Sam007 

Output:

Count of numbers from 1 to 107 is 17

Refer below post for an optimized solution.

Count numbers having 0 as a digit

This article is contributed by Dheeraj Gupta. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

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My Personal Notes

C++

Java

Python3

C#

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