C++ 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: 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
Approach: For large numbers it is difficult to rotate and divide each number by 8. Therefore, ‘divisibility by 8’ property is used which says that a number is divisible by 8 if the last 3 digits of the number is divisible by 8. Here we do not actually rotate the number and check last 8 digits for divisibility, instead we count consecutive sequence of 3 digits (in circular way) which are divisible by 8.
Illustration:
Consider a number 928160
Its rotations are 928160, 092816, 609281,
160928, 816092, 281609.
Now form consecutive sequence of 3-digits from
the original number 928160 as mentioned in the
approach.
3-digit: (9, 2, 8), (2, 8, 1), (8, 1, 6),
(1, 6, 0),(6, 0, 9), (0, 9, 2)
We can observe that the 3-digit number formed by
the these sets, i.e., 928, 281, 816, 160, 609, 092,
are present in the last 3 digits of some rotation.
Thus, checking divisibility of these 3-digit numbers
gives the required number of rotations.
C++
#include <bits/stdc++.h>
using namespace std;
int countRotationsDivBy8(string n)
{
int len = n.length();
int count = 0;
if (len == 1) {
int oneDigit = n[0] - '0' ;
if (oneDigit % 8 == 0)
return 1;
return 0;
}
if (len == 2) {
int first = (n[0] - '0' ) * 10 + (n[1] - '0' );
int second = (n[1] - '0' ) * 10 + (n[0] - '0' );
if (first % 8 == 0)
count++;
if (second % 8 == 0)
count++;
return count;
}
int threeDigit;
for ( int i = 0; i < (len - 2); i++) {
threeDigit = (n[i] - '0' ) * 100 +
(n[i + 1] - '0' ) * 10 +
(n[i + 2] - '0' );
if (threeDigit % 8 == 0)
count++;
}
threeDigit = (n[len - 1] - '0' ) * 100 +
(n[0] - '0' ) * 10 +
(n[1] - '0' );
if (threeDigit % 8 == 0)
count++;
threeDigit = (n[len - 2] - '0' ) * 100 +
(n[len - 1] - '0' ) * 10 +
(n[0] - '0' );
if (threeDigit % 8 == 0)
count++;
return count;
}
int main()
{
string n = "43262488612" ;
cout << "Rotations: "
<< countRotationsDivBy8(n);
return 0;
}
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Output:
Rotations: 4
Time Complexity: O(n), where n is the number of digits in the input number.
Auxiliary Space: O(1)
Please refer complete article on Count rotations divisible by 8 for more details!
Last Updated :
09 Jun, 2022
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