Given a large positive number as string, count all rotations of the given number which are divisible by 4.
Examples:
Input: 8 Output: 1 Input: 20 Output: 1 Rotation: 20 is divisible by 4 02 is not divisible by 4 Input : 13502 Output : 0 No rotation is divisible by 4 Input : 43292816 Output : 5 5 rotations are : 43292816, 16432928, 81643292 92816432, 32928164
For large numbers it is difficult to rotate and divide each number by 4. Therefore, ‘divisibility by 4’ property is used which says that a number is divisible by 4 if the last 2 digits of the number is divisible by 4. Here we do not actually rotate the number and check last 2 digits for divisibility, instead we count consecutive pairs (in circular way) which are divisible by 4.
Illustration:
Consider a number 928160 Its rotations are 928160, 092816, 609281, 160928, 816092, 281609. Now form pairs from the original number 928160 as mentioned in the approach. Pairs: (9,2), (2,8), (8,1), (1,6), (6,0), (0,9) We can observe that the 2-digit number formed by the these pairs, i.e., 92, 28, 81, 16, 60, 09, are present in the last 2 digits of some rotation. Thus, checking divisibility of these pairs gives the required number of rotations. Note: A single digit number can directly be checked for divisibility.
Below is the implementation of the approach.
// C++ program to count all rotation divisible // by 4. #include <bits/stdc++.h> using namespace std;
// Returns count of all rotations divisible // by 4 int countRotations(string n)
{ int len = n.length();
// For single digit number
if (len == 1)
{
int oneDigit = n.at(0)- '0' ;
if (oneDigit%4 == 0)
return 1;
return 0;
}
// At-least 2 digit number (considering all
// pairs)
int twoDigit, count = 0;
for ( int i=0; i<(len-1); i++)
{
twoDigit = (n.at(i)- '0' )*10 + (n.at(i+1)- '0' );
if (twoDigit%4 == 0)
count++;
}
// Considering the number formed by the pair of
// last digit and 1st digit
twoDigit = (n.at(len-1)- '0' )*10 + (n.at(0)- '0' );
if (twoDigit%4 == 0)
count++;
return count;
} //Driver program int main()
{ string n = "4834" ;
cout << "Rotations: " << countRotations(n) << endl;
return 0;
} |
// Java program to count // all rotation divisible // by 4. import java.io.*;
class GFG {
// Returns count of all
// rotations divisible
// by 4
static int countRotations(String n)
{
int len = n.length();
// For single digit number
if (len == 1 )
{
int oneDigit = n.charAt( 0 )- '0' ;
if (oneDigit % 4 == 0 )
return 1 ;
return 0 ;
}
// At-least 2 digit
// number (considering all
// pairs)
int twoDigit, count = 0 ;
for ( int i = 0 ; i < (len- 1 ); i++)
{
twoDigit = (n.charAt(i)- '0' ) * 10 +
(n.charAt(i+ 1 )- '0' );
if (twoDigit% 4 == 0 )
count++;
}
// Considering the number
// formed by the pair of
// last digit and 1st digit
twoDigit = (n.charAt(len- 1 )- '0' ) * 10 +
(n.charAt( 0 )- '0' );
if (twoDigit% 4 == 0 )
count++;
return count;
}
//Driver program
public static void main(String args[])
{
String n = "4834" ;
System.out.println( "Rotations: " +
countRotations(n));
}
} // This code is contributed by Nikita tiwari. |
# Python3 program to count # all rotation divisible # by 4. # Returns count of all # rotations divisible # by 4 def countRotations(n) :
l = len (n)
# For single digit number
if (l = = 1 ) :
oneDigit = ( int )(n[ 0 ])
if (oneDigit % 4 = = 0 ) :
return 1
return 0
# At-least 2 digit number
# (considering all pairs)
count = 0
for i in range ( 0 , l - 1 ) :
twoDigit = ( int )(n[i]) * 10 + ( int )(n[i + 1 ])
if (twoDigit % 4 = = 0 ) :
count = count + 1
# Considering the number
# formed by the pair of
# last digit and 1st digit
twoDigit = ( int )(n[l - 1 ]) * 10 + ( int )(n[ 0 ])
if (twoDigit % 4 = = 0 ) :
count = count + 1
return count
# Driver program n = "4834"
print ( "Rotations: " ,
countRotations(n))
# This code is contributed by Nikita tiwari. |
// C# program to count all rotation // divisible by 4. using System;
class GFG {
// Returns count of all
// rotations divisible
// by 4
static int countRotations(String n)
{
int len = n.Length;
// For single digit number
if (len == 1)
{
int oneDigit = n[0] - '0' ;
if (oneDigit % 4 == 0)
return 1;
return 0;
}
// At-least 2 digit
// number (considering all
// pairs)
int twoDigit, count = 0;
for ( int i = 0; i < (len - 1); i++)
{
twoDigit = (n[i] - '0' ) * 10 +
(n[i + 1] - '0' );
if (twoDigit % 4 == 0)
count++;
}
// Considering the number
// formed by the pair of
// last digit and 1st digit
twoDigit = (n[len - 1] - '0' ) * 10 +
(n[0] - '0' );
if (twoDigit % 4 == 0)
count++;
return count;
}
//Driver program
public static void Main()
{
String n = "4834" ;
Console.Write( "Rotations: " +
countRotations(n));
}
} // This code is contributed by nitin mittal. |
<?php // PHP program to count all // rotation divisible by 4. // Returns count of all // rotations divisible // by 4 function countRotations( $n )
{ $len = strlen ( $n );
// For single digit number
if ( $len == 1)
{
$oneDigit = $n [0] - '0' ;
if ( $oneDigit % 4 == 0)
return 1;
return 0;
}
// At-least 2 digit
// number (considering all
// pairs)
$twoDigit ; $count = 0;
for ( $i = 0; $i < ( $len - 1); $i ++)
{
$twoDigit = ( $n [ $i ] - '0' ) * 10 +
( $n [ $i + 1] - '0' );
if ( $twoDigit % 4 == 0)
$count ++;
}
// Considering the number
// formed by the pair of
// last digit and 1st digit
$twoDigit = ( $n [ $len - 1] - '0' ) * 10 +
( $n [0] - '0' );
if ( $twoDigit % 4 == 0)
$count ++;
return $count ;
} // Driver Code $n = "4834" ;
echo "Rotations: " ,
countRotations( $n );
// This code is contributed by ajit ?> |
<script> // Javascript program to count all // rotation divisible by 4. // Returns count of all // rotations divisible // by 4 function countRotations(n)
{ let len = n.length;
// For single digit number
if (len == 1)
{
let oneDigit = n[0] - '0' ;
if (oneDigit % 4 == 0)
return 1;
return 0;
}
// At-least 2 digit
// number (considering all
// pairs)
let twoDigit;
let count = 0;
for (let i = 0; i < (len - 1); i++)
{
twoDigit = (n[i] - '0' ) * 10 +
(n[i + 1] - '0' );
if (twoDigit % 4 == 0)
count++;
}
// Considering the number
// formed by the pair of
// last digit and 1st digit
twoDigit = (n[len - 1] - '0' ) * 10 +
(n[0] - '0' );
if (twoDigit % 4 == 0)
count++;
return count;
} // Driver Code let n = "4834" ;
document.write( "Rotations: " +
countRotations(n));
// This code is contributed by _saurabh_jaiswal </script> |
Rotations: 2
Time Complexity : O(n) where n is number of digits in input number.
Auxiliary Space: O(1)