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.
Javascript
<script>// Javascript program to count all// rotation divisible by 4.// Returns count of all// rotations divisible// by 4function 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 Codelet 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!
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