Count how many times the given digital clock shows identical digits

Given a generic digital clock, having h number of hours and m number of minutes, the task is to find how many times the clock shows identical time. A specific time is said to be identical if every digit in the hours and minutes is same i.e. the time is of type D:D, D:DD, DD:D or DD:DD. 
Note that the time is written on the digital clock without any leading zeros and the clock shows time between 0 to h - 1 hours and 0 to m - 1 minutes. Few examples of identical times are: 

• 1:1
• 22:22
• 3:33
• 11:1

Examples: 
 

Input: hours = 24, minutes = 60 
Output: 19 
The clock has 24 hours and 60 minutes. 
So the identical times will be: 
Single digit hours and single digit minutes -> 0:0, 1:1, 2:2, ...., 9:9 
Single digit hours and double digit minutes -> 1:11, 2:22, 3:33, 4:44 and 5:55 
Double digit hours and single digit minutes -> 11:1 and 22:2 
Double digit hours and double digit minutes -> 11:11, 22:22 
Total = 10 + 5 + 2 + 2 = 19
Input: hours = 34, minutes = 50 
Output: 20 
 

Approach: As we can see in the explained example, we have to first count the single-digit (of hours) identical times and then double-digit hours. During each of these counts, we need to consider single-digit minutes as well as double-digit minutes. 
There will be two loops. First loop deals with single-digit hours. And the second deals with double-digit hours. In each of the loops, there should be two conditions. First, if the iterator variable is less than total minutes, then increment the counter. Second, if (iterator variable + iterator variable * 10) is less than total minutes, increment the counter. In the end, we will have the total identical times that clock shows.
 

Below is the implementation of the above approach: 
 

19

Time Complexity: O(1)
Auxiliary Space: O(1)