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PHP Program to Count Rotations Divisible by 4

Last Updated : 22 Jul, 2024
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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. 

PHP
<?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);

?>

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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