Problem: Count how many integers from 1 to N contains 0 as a digit.
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 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:
- 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)
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.
Count of numbers from 1 to 107 is 17 Count of numbers from 1 to 1264 is 315
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 firstname.lastname@example.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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