C++ Program to 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 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; } |
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!