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++ 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 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 */
# 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# 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 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.
?>
<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
19
Time Complexity: O(1)
Auxiliary Space: O(1)