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.
C++
// C++ program to count all rotation divisible // by 4. #include <bits/stdc++.h> using namespace std; // Returns count of all rotations divisible // by 4 int countRotations(string n) { int len = n.length(); // For single digit number if (len == 1) { int oneDigit = n.at(0)- '0' ; if (oneDigit%4 == 0) return 1; return 0; } // At-least 2 digit number (considering all // pairs) int twoDigit, count = 0; for ( int i=0; i<(len-1); i++) { twoDigit = (n.at(i)- '0' )*10 + (n.at(i+1)- '0' ); if (twoDigit%4 == 0) count++; } // Considering the number formed by the pair of // last digit and 1st digit twoDigit = (n.at(len-1)- '0' )*10 + (n.at(0)- '0' ); if (twoDigit%4 == 0) count++; return count; } //Driver program int main() { string n = "4834" ; cout << "Rotations: " << countRotations(n) << endl; return 0; } |
Java
// Java program to count // all rotation divisible // by 4. import java.io.*; class GFG { // Returns count of all // rotations divisible // by 4 static int countRotations(String n) { int len = n.length(); // For single digit number if (len == 1 ) { int oneDigit = n.charAt( 0 )- '0' ; if (oneDigit % 4 == 0 ) return 1 ; return 0 ; } // At-least 2 digit // number (considering all // pairs) int twoDigit, count = 0 ; for ( int i = 0 ; i < (len- 1 ); i++) { twoDigit = (n.charAt(i)- '0' ) * 10 + (n.charAt(i+ 1 )- '0' ); if (twoDigit% 4 == 0 ) count++; } // Considering the number // formed by the pair of // last digit and 1st digit twoDigit = (n.charAt(len- 1 )- '0' ) * 10 + (n.charAt( 0 )- '0' ); if (twoDigit% 4 == 0 ) count++; return count; } //Driver program public static void main(String args[]) { String n = "4834" ; System.out.println( "Rotations: " + countRotations(n)); } } // This code is contributed by Nikita tiwari. |
Python3
# Python 3 program to count # all rotation divisible # by 4. # Returns count of all # rotations divisible # by 4 def countRotations(n) : l = len (n) # For single digit number if (l = = 1 ) : oneDigit = ( int )(n[ 0 ]) if (oneDigit % 4 = = 0 ) : return 1 return 0 # At-least 2 digit number # (considering all pairs) count = 0 for i in range ( 0 , l - 1 ) : twoDigit = ( int )(n[i]) * 10 + ( int )(n[i + 1 ]) if (twoDigit % 4 = = 0 ) : count = count + 1 # Considering the number # formed by the pair of # last digit and 1st digit twoDigit = ( int )(n[l - 1 ]) * 10 + ( int )(n[ 0 ]) if (twoDigit % 4 = = 0 ) : count = count + 1 return count # Driver program n = "4834" print ( "Rotations: " , countRotations(n)) # This code is contributed by Nikita tiwari. |
C#
// C# program to count all rotation // divisible by 4. using System; class GFG { // Returns count of all // rotations divisible // by 4 static int countRotations(String n) { int len = n.Length; // For single digit number if (len == 1) { int oneDigit = n[0] - '0' ; if (oneDigit % 4 == 0) return 1; return 0; } // At-least 2 digit // number (considering all // pairs) int twoDigit, count = 0; for ( int 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 program public static void Main() { String n = "4834" ; Console.Write( "Rotations: " + countRotations(n)); } } // This code is contributed by nitin mittal. |
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 ); // This code is contributed by ajit ?> |
Output:
Rotations: 2
Time Complexity : O(n) where n is number of digits in input number.
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