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.
<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> |
Output:
Rotations: 2
Time Complexity : O(n) where n is number of digits in input number.
Auxiliary Space: O(1)
Please refer complete article on Count rotations divisible by 4 for more details!