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 througn 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; 
  
    // Travers 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 througn 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; 
      
        // Travers 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. 

Python 3

# Python 3 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
  
    # Travers 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 througn 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; 
      
        // Travers 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 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

My Personal Notes

C++

Java

Python 3

C#

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