Rearrangement of a number which is also divisible by it

Given a number n, we need to rearrange all its digits such that the new arrangement is divisible by n. Also, the new number should not be equal to x. If no such rearrangement is possible, print -1. Examples:

Input : n = 1035
Output : 3105
The result 3105 is divisible by
given n and has the same set of digits.

Input : n = 1782
Output : m = 7128

Simple Approach : Find all the permutation of given n and then check whether it is divisible by n or not also check that new permutation should not be equal to n. Efficient Approach : Let’s suppose that y is our result then y = m * n, also we know that y must be a rearrangement of digits of n so we can say now restrict m (the multiplier) as per given conditions. 1) y has the same number of digits as n has. So, m must be less than 10. 2) y must not be equal to n. So, m will be greater than 1. So we get the multiplier m in the range [2,9]. So we will find all the possible y and then check that should y has the same digits as n or not. 

C++

// CPP program for finding rearrangement of n
// that is divisible  by n
#include<bits/stdc++.h>
using namespace std;

// perform hashing for given n
void storeDigitCounts(int n, vector<int> &hash)
{
    // perform hashing
    while (n)
    {
        hash[n%10]++;
        n /= 10;
    }
}

// check whether any arrangement exists
int rearrange (int n)
{
    // Create a hash for given number n
    // The hash is of size 10 and stores
    // count of each digit in n.
    vector<int> hash_n(10, 0);
    storeDigitCounts(n, hash_n);

    // check for all possible multipliers
    for (int mult=2; mult<10; mult++)
    {
        int curr = n * mult;

        vector<int> hash_curr(10, 0);
        storeDigitCounts(curr, hash_curr);

        // check hash table for both.
        // Please refer below link for help
        // of equal()
        // https://www.geeksforgeeks.org/stdequal-in-cpp/
        if (equal(hash_n.begin(), hash_n.end(),
                             hash_curr.begin()))
            return curr;
    }
    return -1;
}

// driver program
int main()
{
    int n = 10035;
    cout << rearrange(n);
    return 0;
}

Java

// Java program for finding rearrangement of n
// that is divisible  by n
import java.util.*;

class GFG
{
  // perform hashing for given n
  static int[] storeDigitCounts(int n)
  {
    int arr[] = new int[10];
    for (int i = 0; i < 10; i++)
      arr[i] = 0;
    // perform hashing
    while (n > 0)
    {
      arr[n % 10]++;
      n /= 10;
    }
    return arr;
  }


  // check whether any arrangement exists
  static int rearrange (int n)
  {
    // Create a hash for given number n
    // The hash is of size 10 and stores
    // count of each digit in n.
    int[] hash = storeDigitCounts(n);

    // check for all possible multipliers
    for (int mult=2; mult<10; mult++)
    {
      int curr = n*mult;

      int[] hash_curr = storeDigitCounts(curr);

      // check hash table for both.
      int ind;
      for (ind = 0; ind < 10; ind++)
      {
        if (hash_curr[ind] != hash[ind])
          break;
      }
      if (ind == 10)  return curr;

    }
    return -1;
  }

  // driver program
  public static void main(String[] args)
  {
    int n = 10035;
    System.out.println(rearrange(n));
  }
}

// This code is contributed by phasing17

Python3

# Python3 program for finding rearrangement
# of n that is divisible by n

# Perform hashing for given n
def storeDigitCounts(n, Hash):

    # perform hashing
    while n > 0:

        Hash[n % 10] += 1
        n //= 10

# check whether any arrangement exists
def rearrange(n):

    # Create a hash for given number n
    # The hash is of size 10 and stores
    # count of each digit in n.
    hash_n = [0] * 10
    storeDigitCounts(n, hash_n)

    # check for all possible multipliers
    for mult in range(2, 10):

        curr = n * mult

        hash_curr = [0] * 10
        storeDigitCounts(curr, hash_curr)

        # check hash table for both.
        if hash_n == hash_curr:
            return curr

    return -1

# Driver Code
if __name__ == "__main__":

    n = 10035
    print(rearrange(n))

# This code is contributed by Rituraj Jain

C#

// C# program for finding rearrangement of n
// that is divisible  by n
using System;
using System.Collections.Generic;

class GFG
{

  // perform hashing for given n
  static int[] storeDigitCounts(int n)
  {
    int[] arr = new int[10];
    for (int i = 0; i < 10; i++)
      arr[i] = 0;
    // perform hashing
    while (n > 0) {
      arr[n % 10]++;
      n /= 10;
    }
    return arr;
  }

  // check whether any arrangement exists
  static int rearrange(int n)
  {
    // Create a hash for given number n
    // The hash is of size 10 and stores
    // count of each digit in n.
    int[] hash = storeDigitCounts(n);

    // check for all possible multipliers
    for (int mult = 2; mult < 10; mult++) {
      int curr = n * mult;

      int[] hash_curr = storeDigitCounts(curr);

      // check hash table for both.
      int ind;
      for (ind = 0; ind < 10; ind++) {
        if (hash_curr[ind] != hash[ind])
          break;
      }
      if (ind == 10)
        return curr;
    }
    return -1;
  }

  // driver program
  public static void Main(string[] args)
  {
    int n = 10035;
    Console.WriteLine(rearrange(n));
  }
}

// This code is contributed by phasing17

Javascript

// JavaScript program for finding rearrangement
// of n that is divisible by n

// Perform hashing for given n
function storeDigitCounts(n, Hash)
{
    // perform hashing
    while (n > 0)
    {
        Hash[n % 10] += 1;
        n = Math.floor(n / 10);
    }
    return Hash;
}

// check whether any arrangement exists
function rearrange(n)
{
    // Create a hash for given number n
    // The hash is of size 10 and stores
    // count of each digit in n.
    let hash_n = new Array(10).fill(0);
    hash_n = storeDigitCounts(n, hash_n);

    // check for all possible multipliers
    for (var mult = 2; mult < 10; mult++)
    {
        let curr = n * mult;

        let hash_curr = new Array(10).fill(0);
        hash_curr = storeDigitCounts(curr, hash_curr) ;

        // check hash table for both.
        if (JSON.stringify(hash_curr)==JSON.stringify(hash_n))
            return curr ;
    }

    return -1;
}

// Driver Code
let n = 10035;
console.log(rearrange(n));

// This code is contributed by phasing17

Output:

30105

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