Find the smallest number whose digits multiply to a given number n

Given a number ‘n’, find the smallest number ‘p’ such that if we multiply all digits of ‘p’, we get ‘n’. The result ‘p’ should have minimum two digits.

Examples:

Input:  n = 36

Output: p = 49 

// Note that 4*9 = 36 and 49 is the smallest such number



Input:  n = 100

Output: p = 455

// Note that 4*5*5 = 100 and 455 is the smallest such number



Input: n = 1

Output:p = 11

// Note that 1*1 = 1



Input: n = 13

Output: Not Possible

For a given n, following are the two cases to be considered.
Case 1: n < 10 When n is smaller than n, the output is always n+10. For example for n = 7, output is 17. For n = 9, output is 19.

Case 2: n >= 10 Find all factors of n which are between 2 and 9 (both inclusive). The idea is to start searching from 9 so that the number of digits in result are minimized. For example 9 is preferred over 33 and 8 is preferred over 24.
Store all found factors in an array. The array would contain digits in non-increasing order, so finally print the array in reverse order.

 

 1     public static int findSmallest(int n){

 2         if(n < 10) return 10 + n;

 3         int out = 0;

 4         int pos = 0;

 5         for(int i = 9; i > 1; i --){

 6             while(n % i == 0){

 7                 n /= i;

 8                 out += i * Math.pow(10, pos);

 9                 pos ++;

10             }

11         }

12         return n ==1 ? out : -1;

13     }