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 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. At the end, we will have the total identical times that clock shows.

Below is the implementation of the above approach:

C++

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// 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 < minutes)
            count++;
  
        // Double digit minutes
        if ((i % 10) < minutes)
            count++;
    }
  
    // Return the required count
    return count;
}
  
// Driver code
int main()
{
    int hours = 24;
    int minutes = 60;
    cout << countIdentical(hours, minutes);
  
    return 0;
}

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Java

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// 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
        {
  
            // Single digit minute
            if (i < minutes) 
            {
                count++;
            }
  
            // Double digit minutes
            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;
        System.out.println(countIdentical(hours, minutes));
    }
}
  
/* This code contributed by PrinciRaj1992 */

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Python3

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# 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):
          
        # Single digit minute
        if (i < minutes):
            count += 1
  
        # Double digit minutes
        if ((i % 10) < minutes):
            count += 1
  
        i += 11
  
    # Return the required count
    return count
  
# Driver code
if __name__ == '__main__':
    hours = 24
    minutes = 60
    print(countIdentical(hours, minutes))
  
# This code is contributed by
# Surendra_Gangwar

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C#

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// 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) 
        {
  
            // Single digit minute
            if (i < minutes) 
            {
                count++;
            }
  
            // Double digit minutes
            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;
        Console.WriteLine(countIdentical(hours, minutes));
    }
}
  
// This code has been contributed by 29AjayKumar

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PHP

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<?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)
    {
  
        // Single digit minute
        if ($i < $minutes)
            $count++;
  
        // Double digit minutes
        if (($i % 10) < $minutes)
            $count++;
    }
  
    // Return the required count
    return $count;
}
  
// Driver Code
$hours = 24;
$minutes = 60;
echo countIdentical($hours, $minutes);
  
// This code is contributed by ajit.
?>

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Output:

19


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