Given a number m, find all numbers which have m digits and are a divisor of their right rotation. Right-rotation of a number N is the result of rotating the digits of N one place to the right and wrapping the least significant digit around so that it becomes the most significant digit. For example, the right-rotation of 4356 is 6435.
Examples:
Input: 2 Output: 11 22 33 44 55 66 77 88 99 Input: 6 Output: 102564 111111 128205 142857 153846 179487 205128 222222 230769 333333 444444 555555 666666 777777 888888 999999 128205 satisfies the condition as 128205 * 4 = 512820.
Brute force approach:
The simplest approach is to traverse all the numbers which are greater than or equal to 10m-1 and less than 10m and check if they satisfy the required condition. We can check it in constant time, so the time complexity for the whole process is O(10m), which is feasible for only small values of m.
Below is the implementation of the above approach:
C++
// C++ program to Generating numbers that // are divisor of their right-rotations #include <bits/stdc++.h>using namespace std; // Function to check if N is a // divisor of its right-rotation bool rightRotationDivisor(int N){ int lastDigit = N % 10; int rightRotation = (lastDigit * pow(10 ,int(log10(N)))) + floor(N / 10); return (rightRotation % N == 0);} // Function to generate m-digit // numbers which are divisor of // their right-rotation void generateNumbers(int m){ for (int i=pow(10,(m - 1));i<pow(10 , m);i++) if (rightRotationDivisor(i)) cout<<i<<endl; } // Driver code int main(){int m = 3;generateNumbers(m);} |
Java
// Java program to Generating numbers that // are divisor of their right-rotations public class GFG { // Function to check if N is a // divisor of its right-rotation static boolean rightRotationDivisor(int N) { int lastDigit = N % 10; int rightRotation = (int)(lastDigit * Math.pow(10 ,(int)(Math.log10(N))) + Math.floor(N / 10)); return (rightRotation % N == 0); } // Function to generate m-digit // numbers which are divisor of // their right-rotation static void generateNumbers(int m) { for (int i= (int)Math.pow(10,(m - 1)); i < Math.pow(10 , m);i++) if (rightRotationDivisor(i)) System.out.println(i); } // Driver code public static void main(String args[]) { int m = 3; generateNumbers(m); } // This Code is contributed by ANKITRAI1} |
Python3
# Python program to Generating numbers that are# divisor of their right-rotationsfrom math import log10# Function to check if N is a# divisor of its right-rotationdef rightRotationDivisor(N): lastDigit = N % 10 rightRotation = (lastDigit * 10 ** int(log10(N)) + N // 10) return rightRotation % N == 0# Function to generate m-digit# numbers which are divisor of# their right-rotationdef generateNumbers(m): for i in range(10 ** (m - 1), 10 ** m): if rightRotationDivisor(i): print(i)# Driver codem = 3generateNumbers(m) |
C#
// C# program to Generating numbers that // are divisor of their right-rotations using System;public class GFG{ // Function to check if N is a // divisor of its right-rotation static bool rightRotationDivisor(int N) { int lastDigit = N % 10; int rightRotation = (int)(lastDigit * Math.Pow(10 ,(int)(Math.Log10(N))) + Math.Floor((double)N/10)); return (rightRotation % N == 0); } // Function to generate m-digit // numbers which are divisor of // their right-rotation static void generateNumbers(int m) { for (int i= (int)Math.Pow(10,(m - 1)); i < Math.Pow(10 , m);i++) if (rightRotationDivisor(i)) Console.WriteLine(i); } // Driver code public static void Main() { int m = 3; generateNumbers(m); } } // This code is contributed by 29AjayKumar |
PHP
<?php // PHP program to Generating numbers that // are divisor of their right-rotations // Function to check if N is a // divisor of its right-rotation function rightRotationDivisor($N){ $lastDigit = $N % 10; $rightRotation = ($lastDigit * pow(10 , (int)(log10($N)))) + floor($N / 10); return ($rightRotation % $N == 0);} // Function to generate m-digit // numbers which are divisor of // their right-rotation function generateNumbers($m){ for ($i = pow(10, ($m - 1)); $i < pow(10 , $m); $i++) if (rightRotationDivisor($i)) echo $i . "\n";}// Driver code $m = 3;generateNumbers($m);// This code is contributed // by ChitraNayal?> |
Javascript
<script>// Javascript program to Generating numbers that // are divisor of their right-rotations // Function to check if N is a // divisor of its right-rotation function rightRotationDivisor(N) { let lastDigit = N % 10; let rightRotation = (lastDigit * Math.pow(10 , Math.floor((Math.log10(N)))) + Math.floor(N / 10)); return (rightRotation % N == 0); } // Function to generate m-digit // numbers which are divisor of // their right-rotation function generateNumbers(m) { for (let i= Math.floor(Math.pow(10,(m - 1))); i < Math.floor(Math.pow(10 , m));i++) if (rightRotationDivisor(i)) document.write(i+"<br>"); } // Driver code let m = 3; generateNumbers(m); // This code is contributed by avanitrachhadiya2155</script> |
Output:
111 222 333 444 555 666 777 888 999
Time complexity: O(10m)
Auxiliary Space: O(1)
Efficient approach:
Let dmdm-1..d3d2d1 be the general form of the numbers that we want to generate. Take x = dmdm-1..d3d2 and y = d1. The condition that we want to satisfy is y * 10m – 1 + x = k * (10x + y) where k is a positive integer. Rearranging the terms, we get x = (y * (10m-1 – k)) / (10k – 1). Thus, if we fix the value of y and k, we can get a value of x such that 10x + y is a number we need to generate.
The value of y can range from 1 to 9; observe that we will not have the case y = 0 as that would make the right rotation y * 10m – 1 + x have m – 1 digit, and the required condition will never be met.
We require x to have exactly m – 1 digit, i.e.
We can observe that the lower bound is always less than unity, so we can keep it at 1 since k has to be a positive integer.
We can use these results to obtain a highly efficient solution that runs with constant time complexity, i.e. O(1). An important point to note is that the numbers obtained may not be in a sorted form, so we need to store and explicitly sort them if we wish to obtain the numbers in a sorted fashion.
Below is the implementation of the above approach:
C++
// C++ program to Generating // numbers that are divisor // of their right-rotations#include <bits/stdc++.h>using namespace std;// Function to generate m-digit// numbers which are divisor of// their right-rotationvoid generateNumbers(int m){ vector<int> numbers; int k_max, x; for (int y = 0; y < 10; y++) { k_max = (int)(pow(10, m - 2) * (10 * y + 1)) / (int)(pow(10, m - 1) + y); for (int k = 1; k <= k_max; k++) { x = (int)(y * (pow(10, m - 1) - k)) / (10 * k - 1); if ((int)(y * (pow(10, m - 1) - k)) % (10 * k - 1) == 0) numbers.push_back(10 * x + y); } } sort(numbers.begin(), numbers.end()); for (int i = 0; i < numbers.size(); i++) cout << (numbers[i]) << endl;}// Driver codeint main(){ int m = 3; generateNumbers(m);}// This code is contributed by Chitranayal |
Java
// Java program to Generating numbers that // are divisor of their right-rotations import java.util.*;import java.io.*;class GFG{ // Function to generate m-digit // numbers which are divisor of // their right-rotation static void generateNumbers(int m) { ArrayList<Integer> numbers = new ArrayList<>(); int k_max, x; for (int y = 0; y < 10; y++) { k_max = (int)(Math.pow(10,m-2) * (10 * y + 1)) / (int)(Math.pow(10, m-1) + y); for (int k = 1; k <= k_max; k++) { x = (int)(y*(Math.pow(10,m-1)-k)) / (10 * k -1); if ((int)(y*(Math.pow(10,m-1)-k)) % (10 * k -1) == 0) numbers.add(10 * x + y); } } Collections.sort(numbers); for (int i = 0; i < numbers.size(); i++) System.out.println(numbers.get(i)); } // Driver code public static void main(String args[]) { int m = 3; generateNumbers(m); }}// This code is contributed by rachana soma |
Python 3
# Python program to Generating numbers that # are divisor of their right-rotations from math import log10# Function to generate m-digit# numbers which are divisor of# their right-rotationdef generateNumbers(m): numbers = [] for y in range(1, 10): k_max = ((10 ** (m - 2) * (10 * y + 1)) // (10 ** (m - 1) + y)) for k in range(1, k_max + 1): x = ((y * (10 ** (m - 1) - k)) // (10 * k - 1)) if ((y * (10 ** (m - 1) - k)) % (10 * k - 1) == 0): numbers.append(10 * x + y) for n in sorted(numbers): print(n)# Driver codem = 3generateNumbers(m) |
C#
// C# program to Generating numbers that // are divisor of their right-rotations using System;using System.Collections.Generic;class GFG{// Function to generate m-digit // numbers which are divisor of // their right-rotation static void generateNumbers(int m){ List<int> numbers = new List<int>(); int k_max, x; for (int y = 0; y < 10; y++) { k_max = (int)(Math.Pow(10, m - 2) * (10 * y + 1)) / (int)(Math.Pow(10, m - 1) + y); for (int k = 1; k <= k_max; k++) { x = (int)(y * (Math.Pow(10, m - 1) - k)) / (10 * k - 1); if ((int)(y * (Math.Pow(10, m - 1) - k)) % (10 * k - 1) == 0) numbers.Add(10 * x + y); } } numbers.Sort(); for (int i = 0; i < numbers.Count; i++) Console.WriteLine(numbers[i]);}// Driver codepublic static void Main(String []args){ int m = 3; generateNumbers(m);}}// This code is contributed by PrinciRaj1992 |
Javascript
<script>// Javascript program to Generating numbers that// are divisor of their right-rotations // Function to generate m-digit // numbers which are divisor of // their right-rotation function generateNumbers(m) { let numbers = []; let k_max, x; for (let y = 0; y < 10; y++) { k_max = Math.floor((Math.pow(10,m-2) * (10 * y + 1)) / Math.floor(Math.pow(10, m-1) + y)); for (let k = 1; k <= k_max; k++) { x = Math.floor((y*(Math.pow(10,m-1)-k)) / (10 * k -1)); if (Math.floor((y*(Math.pow(10,m-1)-k)) % (10 * k -1)) == 0) numbers.push(10 * x + y); } } numbers.sort(function(a,b){return a-b;}); for (let i = 0; i < numbers.length; i++) document.write(numbers[i]+"<br>"); } // Driver code let m = 3; generateNumbers(m);// This code is contributed by rag2127</script> |
Output:
111 222 333 444 555 666 777 888 999
Time complexity: O(nlogn)
Auxiliary Space: O(n)
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