# Count numbers having 0 as a digit

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 inprevious post

In 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:

1. Count of 0 digit integers having maximum of d-1 digits.
2. 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 discussedhere.

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:

1. 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)
2. 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 C++ implementation of above idea.

// 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 digits int zeroUpto(int digits) {     // Refer below article for details     // http://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 integer int 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;      // 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)             remaining = remaining*10 + toInt(num[i]);     }      // Final answer is calculated     int ans = zeroUpto(k-1) + (remaining-non_zero);     return ans; }  // Driver program to test the above functions int 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; }

#### 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.