Smallest x such that 1*n, 2*n, … x*n have all digits from 1 to 9

Given a positive number n. We need to find x such that 1*n, 2*n, 3*n…..x*n gives all 10 digits at least once. If no such x is possible print -1.

Examples:

Input : n = 1692
Output : 3
Explanation:
n = 1692, we got the digits- 1, 2, 6, 9
2*n = 3384, we got the digits- 1, 2, 3, 4, 
6, 8, 9.
3*n = 5076, we got the digits- 1, 2, 3, 4, 
5, 6, 7, 8, 9.
At this step we got all the digits at least
once. Therefore our answer is 3.

Input  : 1
Output : 10

Input  : 0
Output :-1

The idea used here is simple. We start from 1 and keep multiplying with n till we do not get all the 10 digits at least once. In order to keep track of all the digits coming at each iteration we use an temporary array of size 10 initially having all zeroes. Whenever we got a digit first time we will initialize its index in array with 1. When all digits are visited once, we are done.

Following is the implementation of it.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to find x such that 1*n, 2*n, 3*n
// ...x * n have all digits from 1 to 9 at least
// once
#include <bits/stdc++.h>
using namespace std;
  
// Returns smallest value x such that 1*n, 2*n, 
// 3*n ...x * n have all digits from 1 to 9 at 
// least once
int smallestX(int n)
{
    // taking temporary array and variable.
    int temp[10] = { 0 };
  
    if (n == 0)
        return -1;
  
    // iterate till we get all the 10 digits 
    // at least once
    int count = 0, x = 0;
    for (x = 1; count < 10; x++) {
        int y = x * n;
          
        // checking all the digits
        while (y) {
            if (temp[y % 10] == false) {
                count++;
                temp[y % 10] = true;
            }
            y /= 10;
        }
    }
    return x - 1;
}
  
// driver function
int main()
{
    int n = 5;
    cout <<smallestX(n);
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to find x such 
// that 1*n, 2*n, 3*n...x * n
// have all digits from 1 to 9
// at least once
import java.io.*;
import java.util.*;
  
class GFG
{
      
// Returns smallest value x 
// such that 1*n, 2*n, 3*n 
// ...x * n have all digits 
// from 1 to 9 at least once
public static int smallestX(int n)
{
    // taking temporary 
    // array and variable.
    int[] temp = new int[10];
    for(int i = 0; i < 10; i++)
    temp[i] = 0;
  
    if (n == 0)
        return -1;
  
    // iterate till we get 
    // all the 10 digits 
    // at least once
    int count = 0, x = 0;
    for (x = 1; count < 10; x++) 
    {
        int y = x * n;
          
        // checking all
        // the digits
        while (y > 0
        {
            if (temp[y % 10] == 0
            {
                count++;
                temp[y % 10] = 1;
            }
            y /= 10;
        }
    }
    return x - 1;
}
  
// Driver Code
public static void main(String args[])
{
    int n = 5;
    System.out.print(smallestX(n));
}
}
  
// This code is contributed
// by Akanksha Rai(Abby_akku)

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 program to find x such 
# that 1*n, 2*n, 3*n ...x * n
# have all digits from 1 to 9 
# at least once
  
# Returns smallest value x such 
# that 1*n, 2*n, 3*n ...x * n 
# have all digits from 1 to 9 
# at least once
def smallestX(n):
    # taking temporary 
    # array and variable.
    temp = [0]*10
  
    if (n == 0):
        return -1
  
    # iterate till we get 
    # all the 10 digits 
    # at least once
    count = 0
    x = 1
    while(count < 10): 
        y = x * n
          
        # checking all
        # the digits
        while (y>0):
            if (temp[y % 10] == 0):
                count+=1
                temp[y % 10] = 1
            y = int(y / 10)
        x+=1
  
    return x - 1
  
  
# Driver code
if __name__=='__main__':
    n = 5
    print(smallestX(n))
  
# This code is contributed 
# by mits

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to find x such 
// that 1*n, 2*n, 3*n...x * n
// have all digits from 1 to 9
// at least once
using System;
  
class GFG
{
      
// Returns smallest value x 
// such that 1*n, 2*n, 3*n 
// ...x * n have all digits 
// from 1 to 9 at least once
public static int smallestX(int n)
{
    // taking temporary 
    // array and variable.
    int[] temp = new int[10];
    for(int i = 0; i < 10; i++)
    temp[i] = 0;
  
    if (n == 0)
        return -1;
  
    // iterate till we get 
    // all the 10 digits 
    // at least once
    int count = 0, x = 0;
    for (x = 1; count < 10; x++) 
    {
        int y = x * n;
          
        // checking all the digits
        while (y > 0) 
        {
            if (temp[y % 10] == 0) 
            {
                count++;
                temp[y % 10] = 1;
            }
            y /= 10;
        }
    }
    return x - 1;
}
  
// dDriver Code
static void Main()
{
    int n = 5;
    Console.Write(smallestX(n));
}
}
  
// This code is contributed by mits

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to find x such 
// that 1*n, 2*n, 3*n ...x * n
// have all digits from 1 to 9 
// at least once
  
// Returns smallest value x such 
// that 1*n, 2*n, 3*n ...x * n 
// have all digits from 1 to 9 
// at least once
function smallestX($n)
{
    // taking temporary 
    // array and variable.
    $temp = array_fill(0, 10, false);
  
    if ($n == 0)
        return -1;
  
    // iterate till we get 
    // all the 10 digits 
    // at least once
    $count = 0;
    $x = 0;
    for ($x = 1; $count < 10; $x++) 
    {
        $y = $x * $n;
          
        // checking all
        // the digits
        while ($y)
        {
            if ($temp[$y % 10] == false) 
            {
                $count++;
                $temp[$y % 10] = true;
            }
            $y = (int)($y / 10);
        }
    }
    return $x - 1;
}
  
// Driver code
$n = 5;
echo smallestX($n);
  
// This code is contributed 
// by mits
?>

chevron_right



Output:

18

This article is contributed by Saloni Gupta. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : Mithun Kumar, Akanksha_Rai



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.