Count numbers having 0 as a digit

Last Updated : 23 Mar, 2023

Count how many integers from 1 to N contain 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# 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

JavaScript

<script>

// JavaScript program to count numbers from 1 to n with 
// 0 as a digit 

// Returns 1 if x has 0, else 0 
function has0(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 = Math.floor(x / 10); 
    } 

    return 0; 
} 

// Returns count of numbers from 1 to n with 0 as digit 
function getCount(n) 
{ 
    // Initialize count of numbers having 0 as digit 
    let count = 0; 

    // Traverse through all numbers and for every number 
    // check if it has 0. 
    for (let i=1; i<=n; i++) 
        count += has0(i); 

    return count; 
} 

// Driver program 

    let n = 107; 
    document.write("Count of numbers from 1" + " to "
        + n + " is " + getCount(n)); 


// This code is contributed by Surbhi Tyagi.

</script>

Output

Count of numbers from 1 to 107 is 17

Time Complexity: O(n log n)
Auxiliary Space: O(1)
Refer below post for an optimized solution.
Count numbers having 0 as a digit

Count of unique digits in a given number N

Dheeraj Gupta

Improve

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Count numbers having 0 as a digit

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