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:
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> |
19
Time Complexity: O(1)
Auxiliary Space: O(1)
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