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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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++

 // 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 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 = { 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 <

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;     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 = *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;     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



Output:

18

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Improved By : Mithun Kumar, Akanksha_Rai

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