Javascript Program to Count rotations divisible by 4
Last Updated :
23 Sep, 2024
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
// 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";
console.log("Rotations: " + countRotations(n));
Complexity Analysis:
- 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!
Similar Reads
Javascript Program to Count rotations divisible by 8 Given a large positive number as string, count all rotations of the given number which are divisible by 8.Examples: Input: 8Output: 1Input: 40Output: 1Rotation: 40 is divisible by 8 04 is not divisible by 8Input : 13502Output : 0No rotation is divisible by 8Input : 43262488612Output : 4Approach: It
3 min read
Javascript Program to Count rotations which are divisible by 10 Given a number N, the task is to count all the rotations of the given number which are divisible by 10.Examples: Input: N = 10203 Output: 2 Explanation: There are 5 rotations possible for the given number. They are: 02031, 20310, 03102, 31020, 10203 Out of these rotations, only 20310 and 31020 are d
2 min read
Count rotations divisible by 4 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 rot
6 min read
Count rotations divisible by 8 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 Ap
10 min read
Javascript Program to Rotate digits of a given number by K Given two integers N and K, the task is to rotate the digits of N by K. If K is a positive integer, left rotate its digits. Otherwise, right rotate its digits. Examples: Input: N = 12345, K = 2Output: 34512 Explanation: Left rotating N(= 12345) by K(= 2) modifies N to 34512. Therefore, the required
2 min read