# Count how many times the given digital clock shows identical digits

• Last Updated : 02 Aug, 2021

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 ``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

 ``

## Javascript

 ``

Output

`19`

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

My Personal Notes arrow_drop_up