Given time in hours, find the time in minutes at which the minute hand and hour hand coincide in the next one hour.
Examples:
Input : 3
Output : (180/11) minutes
Input :7
Output : (420/11) minutes
Approach :
1. Take two variables for hour “h1 and h2” and then find the angle “theta” [theta = (30 * h1)] and then divide it by “11” to find the time in minute(m). We are dividing with 11, because the hands of a clock coincide 11s time in 12 hours and 22 times in 24 hours :
Formula : This can be calculated using the formula for time h1 to h2 means [11m / 2 – 30 (h1)]
this implies [ m = ((30 * h1) * 2) / 11 ] ]
[ m = (theta * 2) / 11 ]
where [theta = (30 * h1) ]
where A and B are hours i.e if given hour is (2, 3) then A = 2 and B = 3 .
C++
#include <bits/stdc++.h>
using namespace std;
void find_time( int h1)
{
int theta = 30 * h1;
cout << "(" << (theta * 2) << "/"
<< "11"
<< ")"
<< " minutes" ;
}
int main() {
int h1 = 3;
find_time(h1);
return 0;
}
|
JAVA
import java.io.*;
class GFG
{
static void find_time( int h1)
{
int theta = 30 * h1;
System.out.println( "(" + theta *
2 + "/" +
" 11 ) minutes" );
}
public static void main (String[] args)
{
int h1 = 3 ;
find_time(h1);
}
}
|
Python3
def find_time(h1):
theta = 30 * h1
print ( "(" , end = "")
print ((theta * 2 ), "/ 11) minutes" )
h1 = 3
find_time(h1)
|
C#
using System;
class GFG
{
static void find_time( int h1)
{
int theta = 30 * h1;
Console.WriteLine( "(" + theta * 2 +
"/" + " 11 )minutes" );
}
public static void Main()
{
int h1 = 3;
find_time(h1);
}
}
|
PHP
<?php
function find_time( $h1 )
{
$theta = 30 * $h1 ;
echo ( "(" . ( $theta * 2) .
"/" . "11" . ")" .
" minutes" );
}
$h1 = 3;
find_time( $h1 );
?>
|
Javascript
<script>
function find_time(h1)
{
theta = 30 * h1;
document.write( "(" + (theta * 2) + "/"
+ "11"
+ ")"
+ " minutes" );
}
h1 = 3;
find_time(h1);
</script>
|
Output : (180/11) minutes
Time Complexity: O(1)
Auxiliary Space: O(1)