Given a large positive number as string, count all rotations of the given number which are divisible by 8.
Examples:
Input: 8 Output: 1 Input: 40 Output: 1 Rotation: 40 is divisible by 8 04 is not divisible by 8 Input : 13502 Output : 0 No rotation is divisible by 8 Input : 43262488612 Output : 4
Approach: For large numbers it is difficult to rotate and divide each number by 8. Therefore, ‘divisibility by 8’ property is used which says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not actually rotate the number and check last 8 digits for divisibility, instead we count consecutive sequence of 3 digits (in circular way) which are divisible by 8.
Illustration:
Consider a number 928160 Its rotations are 928160, 092816, 609281, 160928, 816092, 281609. Now form consecutive sequence of 3-digits from the original number 928160 as mentioned in the approach. 3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6), (1, 6, 0),(6, 0, 9), (0, 9, 2) We can observe that the 3-digit number formed by the these sets, i.e., 928, 281, 816, 160, 609, 092, are present in the last 3 digits of some rotation. Thus, checking divisibility of these 3-digit numbers gives the required number of rotations.
// C++ program to count all rotations divisible // by 8 #include <bits/stdc++.h> using namespace std;
// function to count of all rotations divisible // by 8 int countRotationsDivBy8(string n)
{ int len = n.length();
int count = 0;
// For single digit number
if (len == 1) {
int oneDigit = n[0] - '0' ;
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers (considering all
// pairs)
if (len == 2) {
// first pair
int first = (n[0] - '0' ) * 10 + (n[1] - '0' );
// second pair
int second = (n[1] - '0' ) * 10 + (n[0] - '0' );
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three-digit sequences
int threeDigit;
for ( int i = 0; i < (len - 2); i++) {
threeDigit = (n[i] - '0' ) * 100 +
(n[i + 1] - '0' ) * 10 +
(n[i + 2] - '0' );
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n[len - 1] - '0' ) * 100 +
(n[0] - '0' ) * 10 +
(n[1] - '0' );
if (threeDigit % 8 == 0)
count++;
// Considering the number formed by the last
// two digits and the first digit
threeDigit = (n[len - 2] - '0' ) * 100 +
(n[len - 1] - '0' ) * 10 +
(n[0] - '0' );
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
} // Driver program to test above int main()
{ string n = "43262488612" ;
cout << "Rotations: "
<< countRotationsDivBy8(n);
return 0;
} |
// Java program to count all // rotations divisible by 8 import java.io.*;
class GFG
{ // function to count of all
// rotations divisible by 8
static int countRotationsDivBy8(String n)
{
int len = n.length();
int count = 0 ;
// For single digit number
if (len == 1 ) {
int oneDigit = n.charAt( 0 ) - '0' ;
if (oneDigit % 8 == 0 )
return 1 ;
return 0 ;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2 ) {
// first pair
int first = (n.charAt( 0 ) - '0' ) *
10 + (n.charAt( 1 ) - '0' );
// second pair
int second = (n.charAt( 1 ) - '0' ) *
10 + (n.charAt( 0 ) - '0' );
if (first % 8 == 0 )
count++;
if (second % 8 == 0 )
count++;
return count;
}
// considering all three-digit sequences
int threeDigit;
for ( int i = 0 ; i < (len - 2 ); i++)
{
threeDigit = (n.charAt(i) - '0' ) * 100 +
(n.charAt(i + 1 ) - '0' ) * 10 +
(n.charAt(i + 2 ) - '0' );
if (threeDigit % 8 == 0 )
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n.charAt(len - 1 ) - '0' ) * 100 +
(n.charAt( 0 ) - '0' ) * 10 +
(n.charAt( 1 ) - '0' );
if (threeDigit % 8 == 0 )
count++;
// Considering the number formed by the last
// two digits and the first digit
threeDigit = (n.charAt(len - 2 ) - '0' ) * 100 +
(n.charAt(len - 1 ) - '0' ) * 10 +
(n.charAt( 0 ) - '0' );
if (threeDigit % 8 == 0 )
count++;
// required count of rotations
return count;
}
// Driver program
public static void main (String[] args)
{
String n = "43262488612" ;
System.out.println( "Rotations: "
+countRotationsDivBy8(n));
}
} // This code is contributed by vt_m. |
# Python3 program to count all # rotations divisible by 8 # function to count of all # rotations divisible by 8 def countRotationsDivBy8(n):
l = len (n)
count = 0
# For single digit number
if (l = = 1 ):
oneDigit = int (n[ 0 ])
if (oneDigit % 8 = = 0 ):
return 1
return 0
# For two-digit numbers
# (considering all pairs)
if (l = = 2 ):
# first pair
first = int (n[ 0 ]) * 10 + int (n[ 1 ])
# second pair
second = int (n[ 1 ]) * 10 + int (n[ 0 ])
if (first % 8 = = 0 ):
count + = 1
if (second % 8 = = 0 ):
count + = 1
return count
# considering all
# three-digit sequences
threeDigit = 0
for i in range ( 0 ,(l - 2 )):
threeDigit = ( int (n[i]) * 100 +
int (n[i + 1 ]) * 10 +
int (n[i + 2 ]))
if (threeDigit % 8 = = 0 ):
count + = 1
# Considering the number
# formed by the last digit
# and the first two digits
threeDigit = ( int (n[l - 1 ]) * 100 +
int (n[ 0 ]) * 10 +
int (n[ 1 ]))
if (threeDigit % 8 = = 0 ):
count + = 1
# Considering the number
# formed by the last two
# digits and the first digit
threeDigit = ( int (n[l - 2 ]) * 100 +
int (n[l - 1 ]) * 10 +
int (n[ 0 ]))
if (threeDigit % 8 = = 0 ):
count + = 1
# required count
# of rotations
return count
# Driver Code if __name__ = = '__main__' :
n = "43262488612"
print ( "Rotations:" ,countRotationsDivBy8(n))
# This code is contributed by mits. |
// C# program to count all // rotations divisible by 8 using System;
class GFG {
// function to count of all
// rotations divisible by 8
static int countRotationsDivBy8(String n)
{
int len = n.Length;
int count = 0;
// For single digit number
if (len == 1)
{
int oneDigit = n[0] - '0' ;
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2)
{
// first pair
int first = (n[0] - '0' ) *
10 + (n[1] - '0' );
// second pair
int second = (n[1] - '0' ) *
10 + (n[0] - '0' );
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// considering all three -
// digit sequences
int threeDigit;
for ( int i = 0; i < (len - 2); i++)
{
threeDigit = (n[i] - '0' ) * 100 +
(n[i + 1] - '0' ) * 10 +
(n[i + 2] - '0' );
if (threeDigit % 8 == 0)
count++;
}
// Considering the number formed by the
// last digit and the first two digits
threeDigit = (n[len - 1] - '0' ) * 100 +
(n[0] - '0' ) * 10 +
(n[1] - '0' );
if (threeDigit % 8 == 0)
count++;
// Considering the number formed
// by the last two digits and
// the first digit
threeDigit = (n[len - 2] - '0' ) * 100 +
(n[len - 1] - '0' ) * 10 +
(n[0] - '0' );
if (threeDigit % 8 == 0)
count++;
// required count of rotations
return count;
}
// Driver Code
public static void Main ()
{
String n = "43262488612" ;
Console.Write( "Rotations: "
+countRotationsDivBy8(n));
}
} // This code is contributed by Nitin Mittal. |
<?php // PHP program to count all // rotations divisible by 8 // function to count of all // rotations divisible by 8 function countRotationsDivBy8( $n )
{ $len = strlen ( $n );
$count = 0;
// For single digit number
if ( $len == 1)
{
$oneDigit = $n [0] - '0' ;
if ( $oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if ( $len == 2)
{
// first pair
$first = ( $n [0] - '0' ) * 10 +
( $n [1] - '0' );
// second pair
$second = ( $n [1] - '0' ) * 10 +
( $n [0] - '0' );
if ( $first % 8 == 0)
$count ++;
if ( $second % 8 == 0)
$count ++;
return $count ;
}
// considering all
// three-digit sequences
$threeDigit ;
for ( $i = 0; $i < ( $len - 2); $i ++)
{
$threeDigit = ( $n [ $i ] - '0' ) * 100 +
( $n [ $i + 1] - '0' ) * 10 +
( $n [ $i + 2] - '0' );
if ( $threeDigit % 8 == 0)
$count ++;
}
// Considering the number
// formed by the last digit
// and the first two digits
$threeDigit = ( $n [ $len - 1] - '0' ) * 100 +
( $n [0] - '0' ) * 10 +
( $n [1] - '0' );
if ( $threeDigit % 8 == 0)
$count ++;
// Considering the number
// formed by the last two
// digits and the first digit
$threeDigit = ( $n [ $len - 2] - '0' ) * 100 +
( $n [ $len - 1] - '0' ) * 10 +
( $n [0] - '0' );
if ( $threeDigit % 8 == 0)
$count ++;
// required count
// of rotations
return $count ;
} // Driver Code $n = "43262488612" ;
echo "Rotations: " .
countRotationsDivBy8( $n );
// This code is contributed by mits. ?> |
<script> // Javascript program to count all // rotations divisible by 8 // Function to count of all // rotations divisible by 8 function countRotationsDivBy8(n)
{ let len = n.length;
let count = 0;
// For single digit number
if (len == 1)
{
let oneDigit = n[0] - '0' ;
if (oneDigit % 8 == 0)
return 1;
return 0;
}
// For two-digit numbers
// (considering all pairs)
if (len == 2)
{
// first pair
let first = (n[0] - '0' ) * 10 +
(n[1] - '0' );
// second pair
let second = (n[1] - '0' ) * 10 +
(n[0] - '0' );
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
// Considering all
// three-digit sequences
let threeDigit;
for (let i = 0; i < (len - 2); i++)
{
threeDigit = (n[i] - '0' ) * 100 +
(n[i + 1] - '0' ) * 10 +
(n[i + 2] - '0' );
if (threeDigit % 8 == 0)
count++;
}
// Considering the number
// formed by the last digit
// and the first two digits
threeDigit = (n[len - 1] - '0' ) * 100 +
(n[0] - '0' ) * 10 +
(n[1] - '0' );
if (threeDigit % 8 == 0)
count++;
// Considering the number
// formed by the last two
// digits and the first digit
threeDigit = (n[len - 2] - '0' ) * 100 +
(n[len - 1] - '0' ) * 10 +
(n[0] - '0' );
if (threeDigit % 8 == 0)
count++;
// Required count
// of rotations
return count;
} // Driver Code let n = "43262488612" ;
document.write( "Rotations: " +
countRotationsDivBy8(n));
// This code is contributed by _saurabh_jaiswal </script> |
Output:
Rotations: 4
Time Complexity : O(n), where n is the number of digits in input number.
Auxiliary Space: O(1)