Count how many times the given digital clock shows identical digits

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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:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return the count of
// identical times the clock shows
int countIdentical(int hours, int minutes)
{
 
    // To store the count of identical times
    // Initialized to 1 because of 0:0
    int i, count = 1;
 
    // For single digit hour
    for (i = 1; i <= 9 && i < hours; i++) {
 
        // Single digit minute
        if (i < minutes)
            count++;
 
        // Double digit minutes
        if ((i * 10 + i) < minutes)
            count++;
    }
 
    // For double digit hours
    for (i = 11; i <= 99 && i < hours; i = i + 11) {
 
        // Single digit minute
        if ((i % 10) < minutes)
            count++;
 
        // Double digit minutes
        if (i < minutes)
            count++;
    }
 
    // Return the required count
    return count;
}
 
// Driver code
int main()
{
    int hours = 24;
    int minutes = 60;
   
      // Function Call
    cout << countIdentical(hours, minutes);
 
    return 0;
}

Java

// Java implementation of the above approach
class GFG {
 
    // Function to return the count of
    // identical times the clock shows
    static int countIdentical(int hours, int minutes)
    {
 
        // To store the count of identical times
        // Initialized to 1 because of 0:0
        int i, count = 1;
 
        // For single digit hour
        for (i = 1; i <= 9 && i < hours; i++) {
 
            // Single digit minute
            if (i < minutes) {
                count++;
            }
 
            // Double digit minutes
            if ((i * 10 + i) < minutes) {
                count++;
            }
        }
 
        // For double digit hours
        for (i = 11; i <= 99 && i < hours; i = i + 11) {
 
            // Double digit minutes
            if (i < minutes) {
                count++;
            }
 
            // Single digit minute
            if ((i % 10) < minutes) {
                count++;
            }
        }
 
        // Return the required count
        return count;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int hours = 24;
        int minutes = 60;
       
        // Function Call
        System.out.println(countIdentical(hours, minutes));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python 3 implementation of the approach
 
# Function to return the count of
# identical times the clock shows
 
 
def countIdentical(hours, minutes):
 
    # To store the count of identical times
    # Initialized to 1 because of 0:0
    count = 1
    i = 1
 
    # For single digit hour
    while(i <= 9 and i < hours):
 
        # Single digit minute
        if (i < minutes):
            count += 1
 
        # Double digit minutes
        if ((i * 10 + i) < minutes):
            count += 1
 
        i += 1
 
    # For double digit hours
    i = 11
    while(i <= 99 and i < hours):
 
         # Double digit minutes
        if (i < minutes):
            count += 1
 
        # Single digit minute
        if ((i % 10) < minutes):
            count += 1
 
        i += 11
 
    # Return the required count
    return count
 
 
# Driver code
if __name__ == '__main__':
    hours = 24
    minutes = 60
     
    # Function Call
    print(countIdentical(hours, minutes))
 
# This code is contributed by
# Surendra_Gangwar

C#

// C# implementation of the above approach
using System;
 
class GFG {
 
    // Function to return the count of
    // identical times the clock shows
    static int countIdentical(int hours, int minutes)
    {
 
        // To store the count of identical times
        // Initialized to 1 because of 0:0
        int i, count = 1;
 
        // For single digit hour
        for (i = 1; i <= 9 && i < hours; i++) {
 
            // Single digit minute
            if (i < minutes) {
                count++;
            }
 
            // Double digit minutes
            if ((i * 10 + i) < minutes) {
                count++;
            }
        }
 
        // For double digit hours
        for (i = 11; i <= 99 && i < hours; i = i + 11) {
 
            // Double digit minutes
            if (i < minutes) {
                count++;
            }
 
            // Single digit minute
            if ((i % 10) < minutes) {
                count++;
            }
        }
 
        // Return the required count
        return count;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int hours = 24;
        int minutes = 60;
       
        // Function Call
        Console.WriteLine(countIdentical(hours, minutes));
    }
}
 
// This code has been contributed by 29AjayKumar

PHP

<?php
// PHP implementation of the approach
 
// Function to return the count of
// identical times the clock shows
function countIdentical($hours, $minutes)
{
 
    // To store the count of identical times
    // Initialized to 1 because of 0:0
    $i;
    $count = 1;
 
    // For single digit hour
    for ($i = 1; $i <= 9 && $i < $hours; $i++) 
    {
 
        // Single digit minute
        if ($i < $minutes)
            $count++;
 
        // Double digit minutes
        if (($i * 10 + $i) < $minutes)
            $count++;
    }
 
    // For double digit hours
    for ($i = 11; $i <= 99 &&
                  $i < $hours; $i = $i + 11)
    {
 
         
        // Double digit minutes
        if ($i < $minutes)
            $count++;
 
        // Single digit minute
        if (($i % 10) < $minutes)
            $count++;
    }
 
    // Return the required count
    return $count;
}
 
// Driver Code
$hours = 24;
$minutes = 60;
 
// Function call
echo countIdentical($hours, $minutes);
 
// This code is contributed by ajit.
?>

Javascript

<script>
 
// javascript implementation of the above approach    
// Function to return the count of
// identical times the clock shows
    function countIdentical(hours , minutes) {
 
        // To store the count of identical times
        // Initialized to 1 because of 0:0
        var i, count = 1;
 
        // For single digit hour
        for (i = 1; i <= 9 && i < hours; i++) {
 
            // Single digit minute
            if (i < minutes) {
                count++;
            }
 
            // Double digit minutes
            if ((i * 10 + i) < minutes) {
                count++;
            }
        }
 
        // For var digit hours
        for (i = 11; i <= 99 && i < hours; i = i + 11) {
 
            // Double digit minutes
            if (i < minutes) {
                count++;
            }
 
            // Single digit minute
            if ((i % 10) < minutes) {
                count++;
            }
        }
 
        // Return the required count
        return count;
    }
 
    // Driver code
     
        var hours = 24;
        var minutes = 60;
 
        // Function Call
        document.write(countIdentical(hours, minutes));
 
// This code contributed by Rajput-Ji 
 
</script>

Output

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

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