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

  • Difficulty Level : Easy
  • Last Updated : 28 Apr, 2021

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

 

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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 a temporary array of size 10 initially having all zeroes. Whenever we got a digit the 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++




// 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;
}

Java




// 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)

Python3




# 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

C#




// 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

PHP




<?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
?>

Javascript




<script>
 
// Javascript 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.
    let temp = Array.from({length: 10},
              (_, i) => 0);
    for(let i = 0; i < 10; i++)
    temp[i] = 0;
   
    if (n == 0)
        return -1;
   
    // iterate till we get
    // all the 10 digits
    // at least once
    let count = 0, x = 0;
    for (x = 1; count < 10; x++)
    {
        let 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 program
 
    let n = 5;
    document.write(smallestX(n));
     
</script>

Output:  

18

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