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++
// C++ program to count all rotations divisible// by 8#include <bits/stdc++.h>using namespace std;// function to count of all rotations divisible// by 8int countRotationsDivBy8(string n){ int len = n.length(); int count = 0; // For single digit number if (len == 1) { int oneDigit = n[0] - '0'; if (oneDigit % 8 == 0) return 1; return 0; } // For two-digit numbers (considering all // pairs) if (len == 2) { // first pair int first = (n[0] - '0') * 10 + (n[1] - '0'); // second pair int second = (n[1] - '0') * 10 + (n[0] - '0'); if (first % 8 == 0) count++; if (second % 8 == 0) count++; return count; } // considering all three-digit sequences 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++; } // Considering the number formed by the // last digit and the first two digits threeDigit = (n[len - 1] - '0') * 100 + (n[0] - '0') * 10 + (n[1] - '0'); if (threeDigit % 8 == 0) count++; // Considering the number formed by the last // two digits and the first digit threeDigit = (n[len - 2] - '0') * 100 + (n[len - 1] - '0') * 10 + (n[0] - '0'); if (threeDigit % 8 == 0) count++; // required count of rotations return count;}// Driver program to test aboveint main(){ string n = "43262488612"; cout << "Rotations: " << countRotationsDivBy8(n); return 0;} |
Java
// Java program to count all // rotations divisible by 8import java.io.*;class GFG { // function to count of all // rotations divisible by 8 static int countRotationsDivBy8(String n) { int len = n.length(); int count = 0; // For single digit number if (len == 1) { int oneDigit = n.charAt(0) - '0'; if (oneDigit % 8 == 0) return 1; return 0; } // For two-digit numbers // (considering all pairs) if (len == 2) { // first pair int first = (n.charAt(0) - '0') * 10 + (n.charAt(1) - '0'); // second pair int second = (n.charAt(1) - '0') * 10 + (n.charAt(0) - '0'); if (first % 8 == 0) count++; if (second % 8 == 0) count++; return count; } // considering all three-digit sequences int threeDigit; for (int i = 0; i < (len - 2); i++) { threeDigit = (n.charAt(i) - '0') * 100 + (n.charAt(i + 1) - '0') * 10 + (n.charAt(i + 2) - '0'); if (threeDigit % 8 == 0) count++; } // Considering the number formed by the // last digit and the first two digits threeDigit = (n.charAt(len - 1) - '0') * 100 + (n.charAt(0) - '0') * 10 + (n.charAt(1) - '0'); if (threeDigit % 8 == 0) count++; // Considering the number formed by the last // two digits and the first digit threeDigit = (n.charAt(len - 2) - '0') * 100 + (n.charAt(len - 1) - '0') * 10 + (n.charAt(0) - '0'); if (threeDigit % 8 == 0) count++; // required count of rotations return count; } // Driver program public static void main (String[] args) { String n = "43262488612"; System.out.println( "Rotations: " +countRotationsDivBy8(n)); }}// This code is contributed by vt_m. |
Python3
# Python3 program to count all # rotations divisible by 8# function to count of all # rotations divisible by 8def countRotationsDivBy8(n): l = len(n) count = 0 # For single digit number if (l == 1): oneDigit = int(n[0]) if (oneDigit % 8 == 0): return 1 return 0 # For two-digit numbers # (considering all pairs) if (l == 2): # first pair first = int(n[0]) * 10 + int(n[1]) # second pair second = int(n[1]) * 10 + int(n[0]) if (first % 8 == 0): count+=1 if (second % 8 == 0): count+=1 return count # considering all # three-digit sequences threeDigit=0 for i in range(0,(l - 2)): threeDigit = (int(n[i]) * 100 + int(n[i + 1]) * 10 + int(n[i + 2])) if (threeDigit % 8 == 0): count+=1 # Considering the number # formed by the last digit # and the first two digits threeDigit = (int(n[l - 1]) * 100 + int(n[0]) * 10 + int(n[1])) if (threeDigit % 8 == 0): count+=1 # Considering the number # formed by the last two # digits and the first digit threeDigit = (int(n[l - 2]) * 100 + int(n[l - 1]) * 10 + int(n[0])) if (threeDigit % 8 == 0): count+=1 # required count # of rotations return count# Driver Codeif __name__=='__main__': n = "43262488612" print("Rotations:",countRotationsDivBy8(n))# This code is contributed by mits. |
C#
// C# program to count all // rotations divisible by 8using System;class GFG { // function to count of all // rotations divisible by 8 static int countRotationsDivBy8(String n) { int len = n.Length; int count = 0; // For single digit number if (len == 1) { int oneDigit = n[0] - '0'; if (oneDigit % 8 == 0) return 1; return 0; } // For two-digit numbers // (considering all pairs) if (len == 2) { // first pair int first = (n[0] - '0') * 10 + (n[1] - '0'); // second pair int second = (n[1] - '0') * 10 + (n[0] - '0'); if (first % 8 == 0) count++; if (second % 8 == 0) count++; return count; } // considering all three - // digit sequences 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++; } // Considering the number formed by the // last digit and the first two digits threeDigit = (n[len - 1] - '0') * 100 + (n[0] - '0') * 10 + (n[1] - '0'); if (threeDigit % 8 == 0) count++; // Considering the number formed // by the last two digits and // the first digit threeDigit = (n[len - 2] - '0') * 100 + (n[len - 1] - '0') * 10 + (n[0] - '0'); if (threeDigit % 8 == 0) count++; // required count of rotations return count; } // Driver Code public static void Main () { String n = "43262488612"; Console.Write("Rotations: " +countRotationsDivBy8(n)); }}// This code is contributed by Nitin Mittal. |
PHP
<?php// PHP program to count all // rotations divisible by 8// function to count of all // rotations divisible by 8function countRotationsDivBy8($n){ $len = strlen($n); $count = 0; // For single digit number if ($len == 1) { $oneDigit = $n[0] - '0'; if ($oneDigit % 8 == 0) return 1; return 0; } // For two-digit numbers // (considering all pairs) if ($len == 2) { // first pair $first = ($n[0] - '0') * 10 + ($n[1] - '0'); // second pair $second = ($n[1] - '0') * 10 + ($n[0] - '0'); if ($first % 8 == 0) $count++; if ($second % 8 == 0) $count++; return $count; } // considering all // three-digit sequences $threeDigit; for ($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++; } // Considering the number // formed by the last digit // and the first two digits $threeDigit = ($n[$len - 1] - '0') * 100 + ($n[0] - '0') * 10 + ($n[1] - '0'); if ($threeDigit % 8 == 0) $count++; // Considering the number // formed by the last two // digits and the first digit $threeDigit = ($n[$len - 2] - '0') * 100 + ($n[$len - 1] - '0') * 10 + ($n[0] - '0'); if ($threeDigit % 8 == 0) $count++; // required count // of rotations return $count;}// Driver Code$n = "43262488612";echo "Rotations: " . countRotationsDivBy8($n);// This code is contributed by mits.?> |
Javascript
<script>// Javascript program to count all// rotations divisible by 8// Function to count of all// rotations divisible by 8function countRotationsDivBy8(n){ let len = n.length; let count = 0; // For single digit number if (len == 1) { let oneDigit = n[0] - '0'; if (oneDigit % 8 == 0) return 1; return 0; } // For two-digit numbers // (considering all pairs) if (len == 2) { // first pair let first = (n[0] - '0') * 10 + (n[1] - '0'); // second pair let second = (n[1] - '0') * 10 + (n[0] - '0'); if (first % 8 == 0) count++; if (second % 8 == 0) count++; return count; } // Considering all // three-digit sequences let threeDigit; for(let 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++; } // Considering the number // formed by the last digit // and the first two digits threeDigit = (n[len - 1] - '0') * 100 + (n[0] - '0') * 10 + (n[1] - '0'); if (threeDigit % 8 == 0) count++; // Considering the number // formed by the last two // digits and the first digit threeDigit = (n[len - 2] - '0') * 100 + (n[len - 1] - '0') * 10 + (n[0] - '0'); if (threeDigit % 8 == 0) count++; // Required count // of rotations return count;}// Driver Codelet n = "43262488612";document.write("Rotations: " + countRotationsDivBy8(n));// This code is contributed by _saurabh_jaiswal </script> |
Output:
Rotations: 4
Time Complexity : O(n), where n is the number of digits in input number.
Auxiliary Space: O(1)
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