Count numbers having 0 as a digit

19 Apr, 2021
Problem: Count how many integers from 1 to N contains 0 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.
A naive solution is discussed in previous postIn this post an optimized solution is discussed. Let’s analyze the problem closely.Let the given number has d digits .The required answer can be computed  by computing the following two values:  
Count of 0 digit integers having maximum of d-1 digits.Count of 0 digit integers having exactly d digits (less than/ equal to the given number of course!)
Count of 0 digit integers having maximum of d-1 digits.
Count of 0 digit integers having exactly d digits (less than/ equal to the given number of course!)
Therefore, the solution would be the sum of above two.The first part has already been discussed here.How to find the second part? We can find the total number of integers having d digits (less than equal to given number), which don’t contain any zero.To find this we traverse the number, one digit at a time.We find count of non-negative integers as follows:  
If the number at that place is zero, decrement counter by 1 and break (because we can’t move any further, decrement to assure that the number itself contains a zero)else , multiply the (number-1), with power(9, number of digits to the right to it)
If the number at that place is zero, decrement counter by 1 and break (because we can’t move any further, decrement to assure that the number itself contains a zero)
else , multiply the (number-1), with power(9, number of digits to the right to it)
Let’s illustrate with an example.  
Let the number be n = 123. non_zero = 0
We encounter 1 first, 
 add (1-1)*92  to non_zero (= 0+0)

We encounter 2, 
 add (2-1)*91 to non_zero (= 0+9 = 9)

We encounter 3, 
 add (3-1)*90 to non_zero (=9+3 = 12)
We can observe that non_zero denotes the number of integer consisting of 3 digits (not greater than 123) and don’t contain any zero. i.e., (111, 112, ....., 119, 121, 122, 123) (It is recommended to verify it once)Now, one may ask what’s the point of calculating the count of numbers which don’t have any zeroes?Correct! we’re interested to find the count of integers which have zero.However, we can now easily find that by subtracting non_zero from n after ignoring the most significant place.i.e., In our previous example zero = 23 – non_zero = 23-12 =11 and finally we add the two parts to arrive at the required result!!Below is implementation of above idea.  
C++
Java
Python3
C#
PHP
Javascript
//Modified C++ program to count number from 1 to n with// 0 as a digit.#include <bits/stdc++.h>using namespace std; // Returns count of integers having zero upto given digitsint zeroUpto(int digits){    // Refer below article for details    // https://www.geeksforgeeks.org/count-positive-integers-0-digit/    int first = (pow(10,digits)-1)/9;    int second = (pow(9,digits)-1)/8;    return 9 * (first - second);} // utility function to convert character representation// to integerint toInt(char c){    return int(c)-48;} // counts numbers having zero as digits upto a given// number 'num'int countZero(string num){    // k denoted the number of digits in the number    int k = num.length();     // Calculating the total number having zeros,    // which upto k-1 digits    int total = zeroUpto(k-1);     // Now let us calculate the numbers which don't have    // any zeros. In that k digits upto the given number    int non_zero = 0;    for (int i=0; i<num.length(); i++)    {        // If the number itself contains a zero then        // decrement the counter        if (num[i] == '0')        {            non_zero--;            break;        }         // Adding the number of non zero numbers that        // can be formed        non_zero += (toInt(num[i])-1) * (pow(9,k-1-i));    }     int no = 0, remaining = 0,calculatedUpto=0;     // Calculate the number and the remaining after    // ignoring the most significant digit    for (int i=0; i<num.length(); i++)    {        no = no*10 + (toInt(num[i]));        if (i != 0)            calculatedUpto = calculatedUpto*10 + 9;    }    remaining = no-calculatedUpto;     // Final answer is calculated    // It is calculated by subtracting 9....9 (d-1) times    // from no.    int ans = zeroUpto(k-1) + (remaining-non_zero-1);    return ans;} // Driver program to test the above functionsint main(){    string num = "107";    cout << "Count of numbers from 1" << " to "         << num << " is " << countZero(num) << endl;     num = "1264";    cout << "Count of numbers from 1" << " to "         << num << " is " <<countZero(num) << endl;     return 0;}
//Modified Java program to count number from 1 to n with// 0 as a digit. public class GFG {  // Returns count of integers having zero upto given digitsstatic int zeroUpto(int digits){    // Refer below article for details    // https://www.geeksforgeeks.org/count-positive-integers-0-digit/    int first = (int) ((Math.pow(10,digits)-1)/9);    int second = (int) ((Math.pow(9,digits)-1)/8);    return 9 * (first - second);}  // utility function to convert character representation// to integerstatic int toInt(char c){    return (int)(c)-48;}  // counts numbers having zero as digits upto a given// number 'num'static int countZero(String num){    // k denoted the number of digits in the number    int k = num.length();      // Calculating the total number having zeros,    // which upto k-1 digits    int total = zeroUpto(k-1);      // Now let us calculate the numbers which don't have    // any zeros. In that k digits upto the given number    int non_zero = 0;    for (int i=0; i<num.length(); i++)    {        // If the number itself contains a zero then        // decrement the counter        if (num.charAt(i) == '0')        {            non_zero--;            break;        }          // Adding the number of non zero numbers that        // can be formed        non_zero += (toInt(num.charAt(i))-1) * (Math.pow(9,k-1-i));    }      int no = 0, remaining = 0,calculatedUpto=0;      // Calculate the number and the remaining after    // ignoring the most significant digit    for (int i=0; i<num.length(); i++)    {        no = no*10 + (toInt(num.charAt(i)));        if (i != 0)            calculatedUpto = calculatedUpto*10 + 9;    }    remaining = no-calculatedUpto;      // Final answer is calculated    // It is calculated by subtracting 9....9 (d-1) times    // from no.    int ans = zeroUpto(k-1) + (remaining-non_zero-1);    return ans;}  // Driver program to test the above functions     static public void main(String[] args) {        String num = "107";    System.out.println("Count of numbers from 1" + " to "         + num + " is " + countZero(num));      num = "1264";    System.out.println("Count of numbers from 1" + " to "         + num + " is " +countZero(num));    }} // This code is contributed by 29AjayKumar
# Python3 program to count number from 1 to n# with 0 as a digit. # Returns count of integers having zero# upto given digitsdef zeroUpto(digits):         first = int((pow(10, digits) - 1) / 9);    second = int((pow(9, digits) - 1) / 8);    return 9 * (first - second); # counts numbers having zero as digits# upto a given number 'num'def countZero(num):         # k denoted the number of digits    # in the number    k = len(num);     # Calculating the total number having     # zeros, which upto k-1 digits    total = zeroUpto(k - 1);     # Now let us calculate the numbers which    # don't have any zeros. In that k digits    # upto the given number    non_zero = 0;    for i in range(len(num)):                 # If the number itself contains a zero         # then decrement the counter        if (num[i] == '0'):            non_zero -= 1;            break;         # Adding the number of non zero numbers         # that can be formed        non_zero += (((ord(num[i]) - ord('0')) - 1) *                                (pow(9, k - 1 - i)));     no = 0;    remaining = 0;    calculatedUpto = 0;     # Calculate the number and the remaining    # after ignoring the most significant digit    for i in range(len(num)):        no = no * 10 + (ord(num[i]) - ord('0'));        if (i != 0):            calculatedUpto = calculatedUpto * 10 + 9;         remaining = no - calculatedUpto;     # Final answer is calculated. It is calculated     # by subtracting 9....9 (d-1) times from no.    ans = zeroUpto(k - 1) + (remaining - non_zero - 1);    return ans; # Driver Codenum = "107";print("Count of numbers from 1 to",        num, "is", countZero(num)); num = "1264";print("Count of numbers from 1 to",       num, "is", countZero(num)); # This code is contributed by mits
// Modified C# program to count number from 1 to n with// 0 as a digit.  using System;public class GFG{ // Returns count of integers having zero upto given digitsstatic int zeroUpto(int digits){    // Refer below article for details    // https://www.geeksforgeeks.org/count-positive-integers-0-digit/    int first = (int) ((Math.Pow(10,digits)-1)/9);    int second = (int) ((Math.Pow(9,digits)-1)/8);    return 9 * (first - second);} // utility function to convert character representation// to integerstatic int toInt(char c){    return (int)(c)-48;} // counts numbers having zero as digits upto a given// number 'num'static int countZero(String num){    // k denoted the number of digits in the number    int k = num.Length;     // Calculating the total number having zeros,    // which upto k-1 digits    int total = zeroUpto(k-1);     // Now let us calculate the numbers which don't have    // any zeros. In that k digits upto the given number    int non_zero = 0;    for (int i=0; i<num.Length; i++)    {        // If the number itself contains a zero then        // decrement the counter        if (num[i] == '0')        {            non_zero--;            break;        }         // Adding the number of non zero numbers that        // can be formed        non_zero += (toInt(num[i])-1) * (int)(Math.Pow(9,k-1-i));    }     int no = 0, remaining = 0,calculatedUpto=0;     // Calculate the number and the remaining after    // ignoring the most significant digit    for (int i=0; i<num.Length; i++)    {        no = no*10 + (toInt(num[i]));        if (i != 0)            calculatedUpto = calculatedUpto*10 + 9;    }    remaining = no-calculatedUpto;     // Final answer is calculated    // It is calculated by subtracting 9....9 (d-1) times    // from no.    int ans = zeroUpto(k-1) + (remaining-non_zero-1);    return ans;} // Driver program to test the above functions     static public void Main() {        String num = "107";    Console.WriteLine("Count of numbers from 1" + " to "        + num + " is " + countZero(num));     num = "1264";    Console.WriteLine("Count of numbers from 1" + " to "        + num + " is " +countZero(num));    }} // This code is contributed by 29AjayKumar
<?php// PHP program to count// number from 1 to n// with 0 as a digit. // Returns count of integers// having zero upto given digitsfunction zeroUpto($digits){         $first = (int)((pow(10,                    $digits) - 1) / 9);    $second = (int)((pow(9,                    $digits) - 1) / 8);    return 9 * ($first - $second);}  // counts numbers having// zero as digits upto a// given number 'num'function countZero($num){    // k denoted the number    // of digits in the number    $k = strlen($num);     // Calculating the total    // number having zeros,    // which upto k-1 digits    $total = zeroUpto($k-1);     // Now let us calculate    // the numbers which don't    // have any zeros. In that    // k digits upto the given    // number    $non_zero = 0;    for ($i = 0;         $i < strlen($num); $i++)    {        // If the number itself        // contains a zero then        // decrement the counter        if ($num[$i] == '0')        {            $non_zero--;            break;        }         // Adding the number of        // non zero numbers that        // can be formed        $non_zero += (($num[$i] - '0') - 1) *                      (pow(9, $k - 1 - $i));    }     $no = 0;    $remaining = 0;    $calculatedUpto = 0;     // Calculate the number    // and the remaining after    // ignoring the most    // significant digit    for ($i = 0;         $i < strlen($num); $i++)    {        $no = $no * 10 + ($num[$i] - '0');        if ($i != 0)            $calculatedUpto = $calculatedUpto *                                        10 + 9;    }         $remaining = $no - $calculatedUpto;     // Final answer is calculated    // It is calculated by subtracting    // 9....9 (d-1) times from no.    $ans = zeroUpto($k - 1) +                   ($remaining -                    $non_zero - 1);    return $ans;} // Driver Code$num = "107";echo "Count of numbers from 1 to " .                     $num . " is " .             countZero($num) . "\n"; $num = "1264";echo "Count of numbers from 1 to " .                     $num . " is " .                    countZero($num); // This code is contributed// by mits?>
<script> // Modified javascript program to count number from 1 to n with// 0 as a digit. // Returns count of integers having zero upto given digitsfunction zeroUpto(digits){     // Refer below article for details    // https://www.geeksforgeeks.org/count-positive-integers-0-digit/    var first = parseInt( ((Math.pow(10,digits)-1)/9));    var second = parseInt( ((Math.pow(9,digits)-1)/8));    return 9 * (first - second);}  // utility function to convert character representation// to integerfunction toInt(c){    return parseInt((c.charCodeAt(0))-48);}  // counts numbers having zero as digits upto a given// number 'num'function countZero(num){    // k denoted the number of digits in the number    var k = num.length;      // Calculating the total number having zeros,    // which upto k-1 digits    var total = zeroUpto(k-1);      // Now let us calculate the numbers which don't have    // any zeros. In that k digits upto the given number    var non_zero = 0;    for (i=0; i<num.length; i++)    {        // If the number itself contains a zero then        // decrement the counter        if (num.charAt(i) == '0')        {            non_zero--;            break;        }          // Adding the number of non zero numbers that        // can be formed        non_zero += (toInt(num.charAt(i))-1) * (Math.pow(9,k-1-i));    }      var no = 0, remaining = 0,calculatedUpto=0;      // Calculate the number and the remaining after    // ignoring the most significant digit    for (i=0; i<num.length; i++)    {        no = no*10 + (toInt(num.charAt(i)));        if (i != 0)            calculatedUpto = calculatedUpto*10 + 9;    }    remaining = no-calculatedUpto;      // Final answer is calculated    // It is calculated by subtracting 9....9 (d-1) times    // from no.    var ans = zeroUpto(k-1) + (remaining-non_zero-1);    return ans;}  // Driver program to test the above functionsvar num = "107";document.write("Count of numbers from 1" + " to "     + num + " is " + countZero(num)); var num = "1264";document.write("<br>Count of numbers from 1" + " to "     + num + " is " +countZero(num)); // This code is contributed by shikhasingrajput</script>
Output:  
Count of numbers from 1 to 107 is 17 
Count of numbers from 1 to 1264 is 315
Complexity Analysis:Time Complexity : O(d), where d is no. of digits i.e., O(log(n) Auxiliary Space : O(1) 
This article is contributed by Ashutosh Kumar. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
 
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Mithun Kumar
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shikhasingrajput
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